“Closed” form for $sum frac{1}{n^n}$ The 2019 Stack Overflow Developer Survey Results Are...
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“Closed” form for $sum frac{1}{n^n}$
The 2019 Stack Overflow Developer Survey Results Are InSum of reciprocals of powersWhat is $sum_{k=1}^infty frac{1}{k^k}$?How to evaluate $sum_{n=1}^{infty }frac{1}{n^{n}}$?Evaluation of $sum_{x=1}^{infty}x^{-x}$What does $sum_{n=0}^infty 1/n^n$ converge to?How to compute $sum^ {infty} _{n=1} {frac{1}{n^n}}$?Is there a known evaluation of $sum_{k=1}^{infty} frac{1}{k^k}$?What is the value of $1+frac{1}{2^2} + frac{1}{3^3} + frac{1}{4^4} + cdots$?A closed form for $sum_{k = 1}^{infty} k^{-k}$?The miraculous identity?For integer $k > 1$, is $sum_{i=0}^{infty} 1/k^{2^i}$ transcendental or algebraic, or unknown?How does the series $sum_{n=1}^infty frac{(-1)^n cos(n^2 x)}{n}$ behave?The Basel Problem and Theodorus' SpiralWhy do some series converge and others diverge?Does $sum_{n=1}^{infty}frac{cosleft(frac{npi}{2}right)}{sqrt{n}}$ converge?Sum of reciprocal of primes to the primes.Alternating sum involving primesA closed form of $sum_{n=0}^inftyfrac{n!!}{n!}$Sum of an infinite series: $1 + frac12 - frac13- frac14 + frac15 + frac16 - cdots$?Has the arithmetic mean of the Leibniz series been used for computing $pi$?
$begingroup$
Earlier today, I was talking with my friend about some "cool" infinite series and the value they converge to like the Basel problem, Madhava-Leibniz formula for $pi/4, log 2$ and similar alternating series etc.
One series that popped into our discussion was $sumlimits_{n=1}^{infty} frac{1}{n^n}$.
Proving the convergence of this series is trivial but finding the value to which converges has defied me so far. Mathematica says this series converges to $approx 1.29129$.
I tried Googling about this series and found very little information about this series (which is actually surprising since the series looks cool enough to arise in some context).
We were joking that it should have something to do with $pi,e,phi,gamma$ or at the least it must be a transcendental number :-).
My questions are:
- What does this series converge to?
- Does this series arise in any context and are there interesting trivia to be known about this series?
I am actually slightly puzzled that I have not been able to find much about this series on the Internet. (At least my Google search did not yield any interesting results).
sequences-and-series
edited Nov 27 '11 at 10:07
J. M. is not a mathematician
61.2k5152290
$endgroup$
|
show 2 more comments
$begingroup$
Earlier today, I was talking with my friend about some "cool" infinite series and the value they converge to like the Basel problem, Madhava-Leibniz formula for $pi/4, log 2$ and similar alternating series etc.
One series that popped into our discussion was $sumlimits_{n=1}^{infty} frac{1}{n^n}$.
Proving the convergence of this series is trivial but finding the value to which converges has defied me so far. Mathematica says this series converges to $approx 1.29129$.
I tried Googling about this series and found very little information about this series (which is actually surprising since the series looks cool enough to arise in some context).
We were joking that it should have something to do with $pi,e,phi,gamma$ or at the least it must be a transcendental number :-).
My questions are:
- What does this series converge to?
- Does this series arise in any context and are there interesting trivia to be known about this series?
I am actually slightly puzzled that I have not been able to find much about this series on the Internet. (At least my Google search did not yield any interesting results).
sequences-and-series
edited Nov 27 '11 at 10:07
J. M. is not a mathematician
61.2k5152290
$endgroup$
3
$begingroup$
Did you try Plouffe's Invertor?
$endgroup$
– Aryabhata
Feb 10 '11 at 7:49
$begingroup$
@Moron: Nice website. Through the website, I was only able to get the numerical value up to 1000 digits. pi.lacim.uqam.ca/piDATA/sumnn.txt
$endgroup$
– user17762
Feb 10 '11 at 7:58
2
$begingroup$
I asked a related question on MO: mathoverflow.net/questions/33397/… . It is possible to make "deformed" trigonometric functions by replacing $e^{z}$ with $M(z)$, but so far I haven't found references to them in lists of special functions.
$endgroup$
– graveolensa
Feb 10 '11 at 9:32
$begingroup$
@deoxygerbe: Very Interesting! I was thinking of a function on similar lines and study its behavior. The function actually looks a cool function. It converges for all $z$ and is holomorphic. An interesting function to play around with!
$endgroup$
– user17762
Feb 10 '11 at 9:40
$begingroup$
@Sivaram just emailed you what I wrote up about them before.
$endgroup$
– graveolensa
Feb 10 '11 at 9:47
|
show 2 more comments
$begingroup$
Earlier today, I was talking with my friend about some "cool" infinite series and the value they converge to like the Basel problem, Madhava-Leibniz formula for $pi/4, log 2$ and similar alternating series etc.
One series that popped into our discussion was $sumlimits_{n=1}^{infty} frac{1}{n^n}$.
Proving the convergence of this series is trivial but finding the value to which converges has defied me so far. Mathematica says this series converges to $approx 1.29129$.
I tried Googling about this series and found very little information about this series (which is actually surprising since the series looks cool enough to arise in some context).
We were joking that it should have something to do with $pi,e,phi,gamma$ or at the least it must be a transcendental number :-).
My questions are:
- What does this series converge to?
- Does this series arise in any context and are there interesting trivia to be known about this series?
I am actually slightly puzzled that I have not been able to find much about this series on the Internet. (At least my Google search did not yield any interesting results).
sequences-and-series
edited Nov 27 '11 at 10:07
J. M. is not a mathematician
61.2k5152290
$endgroup$
Earlier today, I was talking with my friend about some "cool" infinite series and the value they converge to like the Basel problem, Madhava-Leibniz formula for $pi/4, log 2$ and similar alternating series etc.
One series that popped into our discussion was $sumlimits_{n=1}^{infty} frac{1}{n^n}$.
Proving the convergence of this series is trivial but finding the value to which converges has defied me so far. Mathematica says this series converges to $approx 1.29129$.
I tried Googling about this series and found very little information about this series (which is actually surprising since the series looks cool enough to arise in some context).
We were joking that it should have something to do with $pi,e,phi,gamma$ or at the least it must be a transcendental number :-).
My questions are:
- What does this series converge to?
- Does this series arise in any context and are there interesting trivia to be known about this series?
I am actually slightly puzzled that I have not been able to find much about this series on the Internet. (At least my Google search did not yield any interesting results).
sequences-and-series
sequences-and-series
edited Nov 27 '11 at 10:07
J. M. is not a mathematician
61.2k5152290
edited Nov 27 '11 at 10:07
J. M. is not a mathematician
61.2k5152290
edited Nov 27 '11 at 10:07
J. M. is not a mathematician
61.2k5152290
edited Nov 27 '11 at 10:07
J. M. is not a mathematician
61.2k5152290
edited Nov 27 '11 at 10:07
J. M. is not a mathematician
61.2k5152290
61.2k5152290
asked Feb 10 '11 at 7:37
user17762
3
$begingroup$
Did you try Plouffe's Invertor?
$endgroup$
– Aryabhata
Feb 10 '11 at 7:49
$begingroup$
@Moron: Nice website. Through the website, I was only able to get the numerical value up to 1000 digits. pi.lacim.uqam.ca/piDATA/sumnn.txt
$endgroup$
– user17762
Feb 10 '11 at 7:58
2
$begingroup$
I asked a related question on MO: mathoverflow.net/questions/33397/… . It is possible to make "deformed" trigonometric functions by replacing $e^{z}$ with $M(z)$, but so far I haven't found references to them in lists of special functions.
$endgroup$
– graveolensa
Feb 10 '11 at 9:32
$begingroup$
@deoxygerbe: Very Interesting! I was thinking of a function on similar lines and study its behavior. The function actually looks a cool function. It converges for all $z$ and is holomorphic. An interesting function to play around with!
$endgroup$
– user17762
Feb 10 '11 at 9:40
$begingroup$
@Sivaram just emailed you what I wrote up about them before.
$endgroup$
– graveolensa
Feb 10 '11 at 9:47
|
show 2 more comments
3
$begingroup$
Did you try Plouffe's Invertor?
$endgroup$
– Aryabhata
Feb 10 '11 at 7:49
$begingroup$
@Moron: Nice website. Through the website, I was only able to get the numerical value up to 1000 digits. pi.lacim.uqam.ca/piDATA/sumnn.txt
$endgroup$
– user17762
Feb 10 '11 at 7:58
2
$begingroup$
I asked a related question on MO: mathoverflow.net/questions/33397/… . It is possible to make "deformed" trigonometric functions by replacing $e^{z}$ with $M(z)$, but so far I haven't found references to them in lists of special functions.
$endgroup$
– graveolensa
Feb 10 '11 at 9:32
$begingroup$
@deoxygerbe: Very Interesting! I was thinking of a function on similar lines and study its behavior. The function actually looks a cool function. It converges for all $z$ and is holomorphic. An interesting function to play around with!
$endgroup$
– user17762
Feb 10 '11 at 9:40
$begingroup$
@Sivaram just emailed you what I wrote up about them before.
$endgroup$
– graveolensa
Feb 10 '11 at 9:47
3
3
$begingroup$
Did you try Plouffe's Invertor?
$endgroup$
– Aryabhata
Feb 10 '11 at 7:49
$begingroup$
Did you try Plouffe's Invertor?
$endgroup$
– Aryabhata
Feb 10 '11 at 7:49
$begingroup$
@Moron: Nice website. Through the website, I was only able to get the numerical value up to 1000 digits. pi.lacim.uqam.ca/piDATA/sumnn.txt
$endgroup$
– user17762
Feb 10 '11 at 7:58
$begingroup$
@Moron: Nice website. Through the website, I was only able to get the numerical value up to 1000 digits. pi.lacim.uqam.ca/piDATA/sumnn.txt
$endgroup$
– user17762
Feb 10 '11 at 7:58
2
2
$begingroup$
I asked a related question on MO: mathoverflow.net/questions/33397/… . It is possible to make "deformed" trigonometric functions by replacing $e^{z}$ with $M(z)$, but so far I haven't found references to them in lists of special functions.
$endgroup$
– graveolensa
Feb 10 '11 at 9:32
$begingroup$
I asked a related question on MO: mathoverflow.net/questions/33397/… . It is possible to make "deformed" trigonometric functions by replacing $e^{z}$ with $M(z)$, but so far I haven't found references to them in lists of special functions.
$endgroup$
– graveolensa
Feb 10 '11 at 9:32
$begingroup$
@deoxygerbe: Very Interesting! I was thinking of a function on similar lines and study its behavior. The function actually looks a cool function. It converges for all $z$ and is holomorphic. An interesting function to play around with!
$endgroup$
– user17762
Feb 10 '11 at 9:40
$begingroup$
@deoxygerbe: Very Interesting! I was thinking of a function on similar lines and study its behavior. The function actually looks a cool function. It converges for all $z$ and is holomorphic. An interesting function to play around with!
$endgroup$
– user17762
Feb 10 '11 at 9:40
$begingroup$
@Sivaram just emailed you what I wrote up about them before.
$endgroup$
– graveolensa
Feb 10 '11 at 9:47
$begingroup$
@Sivaram just emailed you what I wrote up about them before.
$endgroup$
– graveolensa
Feb 10 '11 at 9:47
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Certainly you need to check out Sophomore's Dream.
answered Feb 10 '11 at 7:59
Matthew ConroyMatthew Conroy
10.3k42836
$endgroup$
1
$begingroup$
Nice result!
$endgroup$
– user17762
Feb 10 '11 at 8:01
14
$begingroup$
I wonder thy they call it like that. Very few sophomores might ever come up with such an identity...
$endgroup$
– Mariano Suárez-Álvarez
Feb 10 '11 at 9:09
3
$begingroup$
Right. It's just a clever saying, especially considering that the freshman's dream happens all the time.
$endgroup$
– Mitch
Feb 10 '11 at 15:09
$begingroup$
Wikipedia: The name "sophomore's dream" is in contrast to the name "freshman's dream" which is given to the incorrect identity $(x+y)^n = x^n + y^n$.
$endgroup$
– zapyourtumor
Mar 24 '18 at 3:07
add a comment |
$begingroup$
This series turns out to be equal to the integral of $1/x^x$ from $0$ to $1$. Sorry I am bad with the notation needed for math symbols here; but if you rewrite $x^{-x}$ as the exponential of $xlog x$, you can expand it as a taylor series and integrate term by term; to find it equals your sum.
This doesn't give an explicit value; but I think it's a pretty cool identity.
EDIT: Just putting what you said into symbols, probably for the fun of it.
$$int_0^1 x^{-x} dx=int_0^1 e^{-x log x} dx$$
$$ e^{-x log x} =sum_{k=0}^infty (-1)^k frac{x^klog^kx}{k!}$$
But since
$$int_0^1 x^k log^k x dx =(-1)^k frac{k!}{(k+1)^{k+1}}$$
we get
$$int_0^1 x^{-x} dx= sum_{k=0}^infty frac{1}{(k+1)^{k+1}}=sum_{k=1}^infty frac{1}{k^k}$$
edited Feb 25 '12 at 17:30
Pedro Tamaroff♦
97.6k10153299
answered Feb 10 '11 at 8:04
SteveSteve
61949
$endgroup$
1
$begingroup$
If you want superscripts or subscripts of more than one character, you need to enclose them in {}.
$endgroup$
– Ross Millikan
Feb 10 '11 at 15:01
11
$begingroup$
1969 Putnam exam, problem A4: Show that $int_0^1x^x,dx=sum_{n=1}^{infty}(-1)^{n+1}n^{-n}$.
$endgroup$
– Gerry Myerson
May 11 '11 at 1:55
add a comment |
$begingroup$
Liouville's theorem implies that if your constant is irrational then it is in fact transcendental. Unfortunately, the trivial way of proving irrationality doesn't work. On second thought, the same problem also prevents us from applying Liouville's theorem!
answered Feb 10 '11 at 10:42
Yuval FilmusYuval Filmus
49k472148
$endgroup$
$begingroup$
I am unable to immediately see that if the constant is irrational then it has to be a Liouville number. Could you throw light on the proof?
$endgroup$
– user17762
Feb 11 '11 at 2:57
$begingroup$
@Sivaram: You have a point...
$endgroup$
– Yuval Filmus
Feb 11 '11 at 4:45
3
$begingroup$
To apply Liouville's theorem (or the much stronger Roth version) you need an infinite sequence of rational approximations. The smallest upper bound I get on the denominators of the obvious sequence of rational approximations ($sum_{n=1}^l frac{1}{n^n}$) is $q = prod_{i=1}^n i^i$, whereas the best bound I get for $left| frac{p}{q} - alpha right|$ is $frac{1}{n^n} sum_{k=1}^infty frac{1}{k^k}$. That's huge compared to $1/q^r$ so the approximation theory doesn't apply at all - unless you produce a much better sequence of rationals and bound them very tightly.
$endgroup$
– quanta
May 7 '11 at 21:38
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Certainly you need to check out Sophomore's Dream.
answered Feb 10 '11 at 7:59
Matthew ConroyMatthew Conroy
10.3k42836
$endgroup$
1
$begingroup$
Nice result!
$endgroup$
– user17762
Feb 10 '11 at 8:01
14
$begingroup$
I wonder thy they call it like that. Very few sophomores might ever come up with such an identity...
$endgroup$
– Mariano Suárez-Álvarez
Feb 10 '11 at 9:09
3
$begingroup$
Right. It's just a clever saying, especially considering that the freshman's dream happens all the time.
$endgroup$
– Mitch
Feb 10 '11 at 15:09
$begingroup$
Wikipedia: The name "sophomore's dream" is in contrast to the name "freshman's dream" which is given to the incorrect identity $(x+y)^n = x^n + y^n$.
$endgroup$
– zapyourtumor
Mar 24 '18 at 3:07
add a comment |
$begingroup$
Certainly you need to check out Sophomore's Dream.
answered Feb 10 '11 at 7:59
Matthew ConroyMatthew Conroy
10.3k42836
$endgroup$
1
$begingroup$
Nice result!
$endgroup$
– user17762
Feb 10 '11 at 8:01
14
$begingroup$
I wonder thy they call it like that. Very few sophomores might ever come up with such an identity...
$endgroup$
– Mariano Suárez-Álvarez
Feb 10 '11 at 9:09
3
$begingroup$
Right. It's just a clever saying, especially considering that the freshman's dream happens all the time.
$endgroup$
– Mitch
Feb 10 '11 at 15:09
$begingroup$
Wikipedia: The name "sophomore's dream" is in contrast to the name "freshman's dream" which is given to the incorrect identity $(x+y)^n = x^n + y^n$.
$endgroup$
– zapyourtumor
Mar 24 '18 at 3:07
add a comment |
$begingroup$
Certainly you need to check out Sophomore's Dream.
answered Feb 10 '11 at 7:59
Matthew ConroyMatthew Conroy
10.3k42836
$endgroup$
Certainly you need to check out Sophomore's Dream.
answered Feb 10 '11 at 7:59
Matthew ConroyMatthew Conroy
10.3k42836
answered Feb 10 '11 at 7:59
Matthew ConroyMatthew Conroy
10.3k42836
answered Feb 10 '11 at 7:59
Matthew ConroyMatthew Conroy
10.3k42836
answered Feb 10 '11 at 7:59
Matthew ConroyMatthew Conroy
10.3k42836
10.3k42836
1
$begingroup$
Nice result!
$endgroup$
– user17762
Feb 10 '11 at 8:01
14
$begingroup$
I wonder thy they call it like that. Very few sophomores might ever come up with such an identity...
$endgroup$
– Mariano Suárez-Álvarez
Feb 10 '11 at 9:09
3
$begingroup$
Right. It's just a clever saying, especially considering that the freshman's dream happens all the time.
$endgroup$
– Mitch
Feb 10 '11 at 15:09
$begingroup$
Wikipedia: The name "sophomore's dream" is in contrast to the name "freshman's dream" which is given to the incorrect identity $(x+y)^n = x^n + y^n$.
$endgroup$
– zapyourtumor
Mar 24 '18 at 3:07
add a comment |
1
$begingroup$
Nice result!
$endgroup$
– user17762
Feb 10 '11 at 8:01
14
$begingroup$
I wonder thy they call it like that. Very few sophomores might ever come up with such an identity...
$endgroup$
– Mariano Suárez-Álvarez
Feb 10 '11 at 9:09
3
$begingroup$
Right. It's just a clever saying, especially considering that the freshman's dream happens all the time.
$endgroup$
– Mitch
Feb 10 '11 at 15:09
$begingroup$
Wikipedia: The name "sophomore's dream" is in contrast to the name "freshman's dream" which is given to the incorrect identity $(x+y)^n = x^n + y^n$.
$endgroup$
– zapyourtumor
Mar 24 '18 at 3:07
1
1
$begingroup$
Nice result!
$endgroup$
– user17762
Feb 10 '11 at 8:01
$begingroup$
Nice result!
$endgroup$
– user17762
Feb 10 '11 at 8:01
14
14
$begingroup$
I wonder thy they call it like that. Very few sophomores might ever come up with such an identity...
$endgroup$
– Mariano Suárez-Álvarez
Feb 10 '11 at 9:09
$begingroup$
I wonder thy they call it like that. Very few sophomores might ever come up with such an identity...
$endgroup$
– Mariano Suárez-Álvarez
Feb 10 '11 at 9:09
3
3
$begingroup$
Right. It's just a clever saying, especially considering that the freshman's dream happens all the time.
$endgroup$
– Mitch
Feb 10 '11 at 15:09
$begingroup$
Right. It's just a clever saying, especially considering that the freshman's dream happens all the time.
$endgroup$
– Mitch
Feb 10 '11 at 15:09
$begingroup$
Wikipedia: The name "sophomore's dream" is in contrast to the name "freshman's dream" which is given to the incorrect identity $(x+y)^n = x^n + y^n$.
$endgroup$
– zapyourtumor
Mar 24 '18 at 3:07
$begingroup$
Wikipedia: The name "sophomore's dream" is in contrast to the name "freshman's dream" which is given to the incorrect identity $(x+y)^n = x^n + y^n$.
$endgroup$
– zapyourtumor
Mar 24 '18 at 3:07
add a comment |
$begingroup$
This series turns out to be equal to the integral of $1/x^x$ from $0$ to $1$. Sorry I am bad with the notation needed for math symbols here; but if you rewrite $x^{-x}$ as the exponential of $xlog x$, you can expand it as a taylor series and integrate term by term; to find it equals your sum.
This doesn't give an explicit value; but I think it's a pretty cool identity.
EDIT: Just putting what you said into symbols, probably for the fun of it.
$$int_0^1 x^{-x} dx=int_0^1 e^{-x log x} dx$$
$$ e^{-x log x} =sum_{k=0}^infty (-1)^k frac{x^klog^kx}{k!}$$
But since
$$int_0^1 x^k log^k x dx =(-1)^k frac{k!}{(k+1)^{k+1}}$$
we get
$$int_0^1 x^{-x} dx= sum_{k=0}^infty frac{1}{(k+1)^{k+1}}=sum_{k=1}^infty frac{1}{k^k}$$
edited Feb 25 '12 at 17:30
Pedro Tamaroff♦
97.6k10153299
answered Feb 10 '11 at 8:04
SteveSteve
61949
$endgroup$
1
$begingroup$
If you want superscripts or subscripts of more than one character, you need to enclose them in {}.
$endgroup$
– Ross Millikan
Feb 10 '11 at 15:01
11
$begingroup$
1969 Putnam exam, problem A4: Show that $int_0^1x^x,dx=sum_{n=1}^{infty}(-1)^{n+1}n^{-n}$.
$endgroup$
– Gerry Myerson
May 11 '11 at 1:55
add a comment |
$begingroup$
This series turns out to be equal to the integral of $1/x^x$ from $0$ to $1$. Sorry I am bad with the notation needed for math symbols here; but if you rewrite $x^{-x}$ as the exponential of $xlog x$, you can expand it as a taylor series and integrate term by term; to find it equals your sum.
This doesn't give an explicit value; but I think it's a pretty cool identity.
EDIT: Just putting what you said into symbols, probably for the fun of it.
$$int_0^1 x^{-x} dx=int_0^1 e^{-x log x} dx$$
$$ e^{-x log x} =sum_{k=0}^infty (-1)^k frac{x^klog^kx}{k!}$$
But since
$$int_0^1 x^k log^k x dx =(-1)^k frac{k!}{(k+1)^{k+1}}$$
we get
$$int_0^1 x^{-x} dx= sum_{k=0}^infty frac{1}{(k+1)^{k+1}}=sum_{k=1}^infty frac{1}{k^k}$$
edited Feb 25 '12 at 17:30
Pedro Tamaroff♦
97.6k10153299
answered Feb 10 '11 at 8:04
SteveSteve
61949
$endgroup$
1
$begingroup$
If you want superscripts or subscripts of more than one character, you need to enclose them in {}.
$endgroup$
– Ross Millikan
Feb 10 '11 at 15:01
11
$begingroup$
1969 Putnam exam, problem A4: Show that $int_0^1x^x,dx=sum_{n=1}^{infty}(-1)^{n+1}n^{-n}$.
$endgroup$
– Gerry Myerson
May 11 '11 at 1:55
add a comment |
$begingroup$
This series turns out to be equal to the integral of $1/x^x$ from $0$ to $1$. Sorry I am bad with the notation needed for math symbols here; but if you rewrite $x^{-x}$ as the exponential of $xlog x$, you can expand it as a taylor series and integrate term by term; to find it equals your sum.
This doesn't give an explicit value; but I think it's a pretty cool identity.
EDIT: Just putting what you said into symbols, probably for the fun of it.
$$int_0^1 x^{-x} dx=int_0^1 e^{-x log x} dx$$
$$ e^{-x log x} =sum_{k=0}^infty (-1)^k frac{x^klog^kx}{k!}$$
But since
$$int_0^1 x^k log^k x dx =(-1)^k frac{k!}{(k+1)^{k+1}}$$
we get
$$int_0^1 x^{-x} dx= sum_{k=0}^infty frac{1}{(k+1)^{k+1}}=sum_{k=1}^infty frac{1}{k^k}$$
edited Feb 25 '12 at 17:30
Pedro Tamaroff♦
97.6k10153299
answered Feb 10 '11 at 8:04
SteveSteve
61949
$endgroup$
This series turns out to be equal to the integral of $1/x^x$ from $0$ to $1$. Sorry I am bad with the notation needed for math symbols here; but if you rewrite $x^{-x}$ as the exponential of $xlog x$, you can expand it as a taylor series and integrate term by term; to find it equals your sum.
This doesn't give an explicit value; but I think it's a pretty cool identity.
EDIT: Just putting what you said into symbols, probably for the fun of it.
$$int_0^1 x^{-x} dx=int_0^1 e^{-x log x} dx$$
$$ e^{-x log x} =sum_{k=0}^infty (-1)^k frac{x^klog^kx}{k!}$$
But since
$$int_0^1 x^k log^k x dx =(-1)^k frac{k!}{(k+1)^{k+1}}$$
we get
$$int_0^1 x^{-x} dx= sum_{k=0}^infty frac{1}{(k+1)^{k+1}}=sum_{k=1}^infty frac{1}{k^k}$$
edited Feb 25 '12 at 17:30
Pedro Tamaroff♦
97.6k10153299
answered Feb 10 '11 at 8:04
SteveSteve
61949
edited Feb 25 '12 at 17:30
Pedro Tamaroff♦
97.6k10153299
edited Feb 25 '12 at 17:30
Pedro Tamaroff♦
97.6k10153299
edited Feb 25 '12 at 17:30
Pedro Tamaroff♦
97.6k10153299
97.6k10153299
answered Feb 10 '11 at 8:04
SteveSteve
61949
answered Feb 10 '11 at 8:04
SteveSteve
61949
answered Feb 10 '11 at 8:04
SteveSteve
61949
61949
1
$begingroup$
If you want superscripts or subscripts of more than one character, you need to enclose them in {}.
$endgroup$
– Ross Millikan
Feb 10 '11 at 15:01
11
$begingroup$
1969 Putnam exam, problem A4: Show that $int_0^1x^x,dx=sum_{n=1}^{infty}(-1)^{n+1}n^{-n}$.
$endgroup$
– Gerry Myerson
May 11 '11 at 1:55
add a comment |
1
$begingroup$
If you want superscripts or subscripts of more than one character, you need to enclose them in {}.
$endgroup$
– Ross Millikan
Feb 10 '11 at 15:01
11
$begingroup$
1969 Putnam exam, problem A4: Show that $int_0^1x^x,dx=sum_{n=1}^{infty}(-1)^{n+1}n^{-n}$.
$endgroup$
– Gerry Myerson
May 11 '11 at 1:55
1
1
$begingroup$
If you want superscripts or subscripts of more than one character, you need to enclose them in {}.
$endgroup$
– Ross Millikan
Feb 10 '11 at 15:01
$begingroup$
If you want superscripts or subscripts of more than one character, you need to enclose them in {}.
$endgroup$
– Ross Millikan
Feb 10 '11 at 15:01
11
11
$begingroup$
1969 Putnam exam, problem A4: Show that $int_0^1x^x,dx=sum_{n=1}^{infty}(-1)^{n+1}n^{-n}$.
$endgroup$
– Gerry Myerson
May 11 '11 at 1:55
$begingroup$
1969 Putnam exam, problem A4: Show that $int_0^1x^x,dx=sum_{n=1}^{infty}(-1)^{n+1}n^{-n}$.
$endgroup$
– Gerry Myerson
May 11 '11 at 1:55
add a comment |
$begingroup$
Liouville's theorem implies that if your constant is irrational then it is in fact transcendental. Unfortunately, the trivial way of proving irrationality doesn't work. On second thought, the same problem also prevents us from applying Liouville's theorem!
answered Feb 10 '11 at 10:42
Yuval FilmusYuval Filmus
49k472148
$endgroup$
$begingroup$
I am unable to immediately see that if the constant is irrational then it has to be a Liouville number. Could you throw light on the proof?
$endgroup$
– user17762
Feb 11 '11 at 2:57
$begingroup$
@Sivaram: You have a point...
$endgroup$
– Yuval Filmus
Feb 11 '11 at 4:45
3
$begingroup$
To apply Liouville's theorem (or the much stronger Roth version) you need an infinite sequence of rational approximations. The smallest upper bound I get on the denominators of the obvious sequence of rational approximations ($sum_{n=1}^l frac{1}{n^n}$) is $q = prod_{i=1}^n i^i$, whereas the best bound I get for $left| frac{p}{q} - alpha right|$ is $frac{1}{n^n} sum_{k=1}^infty frac{1}{k^k}$. That's huge compared to $1/q^r$ so the approximation theory doesn't apply at all - unless you produce a much better sequence of rationals and bound them very tightly.
$endgroup$
– quanta
May 7 '11 at 21:38
add a comment |
$begingroup$
Liouville's theorem implies that if your constant is irrational then it is in fact transcendental. Unfortunately, the trivial way of proving irrationality doesn't work. On second thought, the same problem also prevents us from applying Liouville's theorem!
answered Feb 10 '11 at 10:42
Yuval FilmusYuval Filmus
49k472148
$endgroup$
$begingroup$
I am unable to immediately see that if the constant is irrational then it has to be a Liouville number. Could you throw light on the proof?
$endgroup$
– user17762
Feb 11 '11 at 2:57
$begingroup$
@Sivaram: You have a point...
$endgroup$
– Yuval Filmus
Feb 11 '11 at 4:45
3
$begingroup$
To apply Liouville's theorem (or the much stronger Roth version) you need an infinite sequence of rational approximations. The smallest upper bound I get on the denominators of the obvious sequence of rational approximations ($sum_{n=1}^l frac{1}{n^n}$) is $q = prod_{i=1}^n i^i$, whereas the best bound I get for $left| frac{p}{q} - alpha right|$ is $frac{1}{n^n} sum_{k=1}^infty frac{1}{k^k}$. That's huge compared to $1/q^r$ so the approximation theory doesn't apply at all - unless you produce a much better sequence of rationals and bound them very tightly.
$endgroup$
– quanta
May 7 '11 at 21:38
add a comment |
$begingroup$
Liouville's theorem implies that if your constant is irrational then it is in fact transcendental. Unfortunately, the trivial way of proving irrationality doesn't work. On second thought, the same problem also prevents us from applying Liouville's theorem!
answered Feb 10 '11 at 10:42
Yuval FilmusYuval Filmus
49k472148
$endgroup$
Liouville's theorem implies that if your constant is irrational then it is in fact transcendental. Unfortunately, the trivial way of proving irrationality doesn't work. On second thought, the same problem also prevents us from applying Liouville's theorem!
answered Feb 10 '11 at 10:42
Yuval FilmusYuval Filmus
49k472148
edited Feb 11 '11 at 4:44
answered Feb 10 '11 at 10:42
Yuval FilmusYuval Filmus
49k472148
answered Feb 10 '11 at 10:42
Yuval FilmusYuval Filmus
49k472148
answered Feb 10 '11 at 10:42
Yuval FilmusYuval Filmus
49k472148
49k472148
$begingroup$
I am unable to immediately see that if the constant is irrational then it has to be a Liouville number. Could you throw light on the proof?
$endgroup$
– user17762
Feb 11 '11 at 2:57
$begingroup$
@Sivaram: You have a point...
$endgroup$
– Yuval Filmus
Feb 11 '11 at 4:45
3
$begingroup$
To apply Liouville's theorem (or the much stronger Roth version) you need an infinite sequence of rational approximations. The smallest upper bound I get on the denominators of the obvious sequence of rational approximations ($sum_{n=1}^l frac{1}{n^n}$) is $q = prod_{i=1}^n i^i$, whereas the best bound I get for $left| frac{p}{q} - alpha right|$ is $frac{1}{n^n} sum_{k=1}^infty frac{1}{k^k}$. That's huge compared to $1/q^r$ so the approximation theory doesn't apply at all - unless you produce a much better sequence of rationals and bound them very tightly.
$endgroup$
– quanta
May 7 '11 at 21:38
add a comment |
$begingroup$
I am unable to immediately see that if the constant is irrational then it has to be a Liouville number. Could you throw light on the proof?
$endgroup$
– user17762
Feb 11 '11 at 2:57
$begingroup$
@Sivaram: You have a point...
$endgroup$
– Yuval Filmus
Feb 11 '11 at 4:45
3
$begingroup$
To apply Liouville's theorem (or the much stronger Roth version) you need an infinite sequence of rational approximations. The smallest upper bound I get on the denominators of the obvious sequence of rational approximations ($sum_{n=1}^l frac{1}{n^n}$) is $q = prod_{i=1}^n i^i$, whereas the best bound I get for $left| frac{p}{q} - alpha right|$ is $frac{1}{n^n} sum_{k=1}^infty frac{1}{k^k}$. That's huge compared to $1/q^r$ so the approximation theory doesn't apply at all - unless you produce a much better sequence of rationals and bound them very tightly.
$endgroup$
– quanta
May 7 '11 at 21:38
$begingroup$
I am unable to immediately see that if the constant is irrational then it has to be a Liouville number. Could you throw light on the proof?
$endgroup$
– user17762
Feb 11 '11 at 2:57
$begingroup$
I am unable to immediately see that if the constant is irrational then it has to be a Liouville number. Could you throw light on the proof?
$endgroup$
– user17762
Feb 11 '11 at 2:57
$begingroup$
@Sivaram: You have a point...
$endgroup$
– Yuval Filmus
Feb 11 '11 at 4:45
$begingroup$
@Sivaram: You have a point...
$endgroup$
– Yuval Filmus
Feb 11 '11 at 4:45
3
3
$begingroup$
To apply Liouville's theorem (or the much stronger Roth version) you need an infinite sequence of rational approximations. The smallest upper bound I get on the denominators of the obvious sequence of rational approximations ($sum_{n=1}^l frac{1}{n^n}$) is $q = prod_{i=1}^n i^i$, whereas the best bound I get for $left| frac{p}{q} - alpha right|$ is $frac{1}{n^n} sum_{k=1}^infty frac{1}{k^k}$. That's huge compared to $1/q^r$ so the approximation theory doesn't apply at all - unless you produce a much better sequence of rationals and bound them very tightly.
$endgroup$
– quanta
May 7 '11 at 21:38
$begingroup$
To apply Liouville's theorem (or the much stronger Roth version) you need an infinite sequence of rational approximations. The smallest upper bound I get on the denominators of the obvious sequence of rational approximations ($sum_{n=1}^l frac{1}{n^n}$) is $q = prod_{i=1}^n i^i$, whereas the best bound I get for $left| frac{p}{q} - alpha right|$ is $frac{1}{n^n} sum_{k=1}^infty frac{1}{k^k}$. That's huge compared to $1/q^r$ so the approximation theory doesn't apply at all - unless you produce a much better sequence of rationals and bound them very tightly.
$endgroup$
– quanta
May 7 '11 at 21:38
add a comment |
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$begingroup$
Did you try Plouffe's Invertor?
$endgroup$
– Aryabhata
Feb 10 '11 at 7:49
$begingroup$
@Moron: Nice website. Through the website, I was only able to get the numerical value up to 1000 digits. pi.lacim.uqam.ca/piDATA/sumnn.txt
$endgroup$
– user17762
Feb 10 '11 at 7:58
2
$begingroup$
I asked a related question on MO: mathoverflow.net/questions/33397/… . It is possible to make "deformed" trigonometric functions by replacing $e^{z}$ with $M(z)$, but so far I haven't found references to them in lists of special functions.
$endgroup$
– graveolensa
Feb 10 '11 at 9:32
$begingroup$
@deoxygerbe: Very Interesting! I was thinking of a function on similar lines and study its behavior. The function actually looks a cool function. It converges for all $z$ and is holomorphic. An interesting function to play around with!
$endgroup$
– user17762
Feb 10 '11 at 9:40
$begingroup$
@Sivaram just emailed you what I wrote up about them before.
$endgroup$
– graveolensa
Feb 10 '11 at 9:47