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A Double sum with something weird between the summations [closed]

3

2

$begingroup$

$$sum_{k=1}^{infty} Bigg[ dfrac{(-1)^{k-1}}{k} sum_{n=0}^{infty} dfrac{300}{k cdot 2^n +5}Bigg]= ?$$
I've never seen such a complex summation before. Can anyone help me? It would also be extremely helpful if anyone could tell me if they've seen this type of summation, and where to find similar problems. Has it appeared in any mathematics competitions, or are there similar questions for me to attempt?

Thanks in advance!

*note: This is Problem 29 from the Chen Jingrun's Cup Secondary School Mathematics Competition 2018. The answer is 137.

summation

share|cite|improve this question

edited Mar 24 at 5:15

NoChance

3,76621321

asked Mar 23 at 13:02

Lee LaindingoldLee Laindingold

162

$endgroup$

closed as off-topic by mrtaurho, Jack D'Aurizio, Lord Shark the Unknown, Alex Provost, José Carlos Santos Mar 24 at 17:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Jack D'Aurizio, Alex Provost, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. This question is likely to be closed if you don't add more context. Where does the problem come form? Are you trying to evaluate the sum or to prove it converges? If the former, do you have any reason to be that exact evaluation is possible? Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
  $endgroup$
  – saulspatz
  Mar 23 at 13:12
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance $n$ is the index of summation of the inner sum.
  $endgroup$
  – saulspatz
  Mar 23 at 13:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Surprising problem and result !
  $endgroup$
  – Claude Leibovici
  Mar 23 at 15:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $$300int_{0}^{1}sum_{ngeq 0} x^4log(1+x^{2^n}),dx =300int_{0}^{1} -x^4log(1-x),dx=color{red}{137}.$$
  $endgroup$
  – Jack D'Aurizio
  Mar 23 at 20:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance It's $kcdot 2^n + 5$ in the denominator, not $k(2^n + 5)$
  $endgroup$
  – eyeballfrog
  Mar 24 at 5:22
  
  
  

  
  
  
  

 | 
show 2 more comments

3

2

$begingroup$

$$sum_{k=1}^{infty} Bigg[ dfrac{(-1)^{k-1}}{k} sum_{n=0}^{infty} dfrac{300}{k cdot 2^n +5}Bigg]= ?$$
I've never seen such a complex summation before. Can anyone help me? It would also be extremely helpful if anyone could tell me if they've seen this type of summation, and where to find similar problems. Has it appeared in any mathematics competitions, or are there similar questions for me to attempt?

Thanks in advance!

*note: This is Problem 29 from the Chen Jingrun's Cup Secondary School Mathematics Competition 2018. The answer is 137.

summation

share|cite|improve this question

edited Mar 24 at 5:15

NoChance

3,76621321

asked Mar 23 at 13:02

Lee LaindingoldLee Laindingold

162

$endgroup$

closed as off-topic by mrtaurho, Jack D'Aurizio, Lord Shark the Unknown, Alex Provost, José Carlos Santos Mar 24 at 17:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Jack D'Aurizio, Alex Provost, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. This question is likely to be closed if you don't add more context. Where does the problem come form? Are you trying to evaluate the sum or to prove it converges? If the former, do you have any reason to be that exact evaluation is possible? Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
  $endgroup$
  – saulspatz
  Mar 23 at 13:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance $n$ is the index of summation of the inner sum.
  $endgroup$
  – saulspatz
  Mar 23 at 13:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Surprising problem and result !
  $endgroup$
  – Claude Leibovici
  Mar 23 at 15:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $$300int_{0}^{1}sum_{ngeq 0} x^4log(1+x^{2^n}),dx =300int_{0}^{1} -x^4log(1-x),dx=color{red}{137}.$$
  $endgroup$
  – Jack D'Aurizio
  Mar 23 at 20:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance It's $kcdot 2^n + 5$ in the denominator, not $k(2^n + 5)$
  $endgroup$
  – eyeballfrog
  Mar 24 at 5:22
  
  
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

$$sum_{k=1}^{infty} Bigg[ dfrac{(-1)^{k-1}}{k} sum_{n=0}^{infty} dfrac{300}{k cdot 2^n +5}Bigg]= ?$$
I've never seen such a complex summation before. Can anyone help me? It would also be extremely helpful if anyone could tell me if they've seen this type of summation, and where to find similar problems. Has it appeared in any mathematics competitions, or are there similar questions for me to attempt?

Thanks in advance!

*note: This is Problem 29 from the Chen Jingrun's Cup Secondary School Mathematics Competition 2018. The answer is 137.

summation

share|cite|improve this question

edited Mar 24 at 5:15

NoChance

3,76621321

asked Mar 23 at 13:02

Lee LaindingoldLee Laindingold

162

$endgroup$

share|cite|improve this question

edited Mar 24 at 5:15

NoChance

3,76621321

asked Mar 23 at 13:02

Lee LaindingoldLee Laindingold

162

share|cite|improve this question

edited Mar 24 at 5:15

NoChance

3,76621321

asked Mar 23 at 13:02

Lee LaindingoldLee Laindingold

162

edited Mar 24 at 5:15

NoChance

3,76621321

edited Mar 24 at 5:15

NoChance

3,76621321

asked Mar 23 at 13:02

Lee LaindingoldLee Laindingold

162

asked Mar 23 at 13:02

Lee LaindingoldLee Laindingold

162

1 Answer
1

active

oldest

votes

1

$begingroup$

This becomes easier by converting to an integral to get that weird linear term out of the denominator. Jack gave the quick version in the comments, but here's the more detailed explanation.

First, introduce an integral to convert from a zeta-like series to a power series:
begin{multline}
sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^infty frac{300}{2^nk+5} = 300sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^inftyint_0^1 x^{2^nk+4}dx
\= 300int_0^1x^4left[sum_{n=0}^inftysum_{k=1}^infty (-1)^{k-1}frac{x^{2^nk}}{k}right] dx = 300int_0^1x^4left[sum_{n=0}^inftylnleft(1+x^{2^n}right)right] dx,
end{multline}
where we used the series expansion $ln(1+z) = sum_{k=0}^infty (-1)^{k-1}z^k/k$ to do the $k$ summation.

Next, evaluate the $n$ series using
$$
sum_{n=0}^infty ln(1+x^{2^n}) = lnleft[prod_{n=0}^infty left(1+x^{2^n}right)right] = lnleft[sum_{m=0}^infty x^mright] = lnleft(frac{1}{1-x}right) = -ln(1-x).
$$
The identity with the infinite product can be seen by considering expanding each exponent $m$ as sums of powers of $2$. So now we've got our sum in terms of a nice integral,
$$
sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^infty frac{300}{2^nk+5} = -300int_0^1x^4ln(1-x)dx
$$
Last step is to use the identity $int_0^1 x^nln(x)dx =-(n+1)^{-2}$ to get
$$
-300int_0^1x^4ln(1-x)dx = -300int_0^1(1-x)^4ln(x)dx = 300left(1 - frac{4}{4} + frac{6}{9}-frac{4}{16}+frac{1}{25}right) = 137
$$
So, in conclusion,
$$
sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^infty frac{300}{2^nk+5} =137
$$

share|cite|improve this answer

answered Mar 24 at 6:26

eyeballfrogeyeballfrog

7,212633

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks so much! It really helped!
  $endgroup$
  – Lee Laindingold
  Mar 24 at 8:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually,this question appeared in a secondary school maths competition so I doubt any calculus is involved..................but at least i found a way of solving it, thanks!It would also be helpful if anyone could provide a more secondary school solution haha
  $endgroup$
  – Lee Laindingold
  Mar 24 at 8:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LeeLaindingold Infinite series are also calculus, and I believe they are usually taught after elementary integrals.
  $endgroup$
  – eyeballfrog
  Mar 24 at 8:12
  

  
  
  
  

add a comment | 

1

$begingroup$

This becomes easier by converting to an integral to get that weird linear term out of the denominator. Jack gave the quick version in the comments, but here's the more detailed explanation.

First, introduce an integral to convert from a zeta-like series to a power series:
begin{multline}
sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^infty frac{300}{2^nk+5} = 300sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^inftyint_0^1 x^{2^nk+4}dx
\= 300int_0^1x^4left[sum_{n=0}^inftysum_{k=1}^infty (-1)^{k-1}frac{x^{2^nk}}{k}right] dx = 300int_0^1x^4left[sum_{n=0}^inftylnleft(1+x^{2^n}right)right] dx,
end{multline}
where we used the series expansion $ln(1+z) = sum_{k=0}^infty (-1)^{k-1}z^k/k$ to do the $k$ summation.

Next, evaluate the $n$ series using
$$
sum_{n=0}^infty ln(1+x^{2^n}) = lnleft[prod_{n=0}^infty left(1+x^{2^n}right)right] = lnleft[sum_{m=0}^infty x^mright] = lnleft(frac{1}{1-x}right) = -ln(1-x).
$$
The identity with the infinite product can be seen by considering expanding each exponent $m$ as sums of powers of $2$. So now we've got our sum in terms of a nice integral,
$$
sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^infty frac{300}{2^nk+5} = -300int_0^1x^4ln(1-x)dx
$$
Last step is to use the identity $int_0^1 x^nln(x)dx =-(n+1)^{-2}$ to get
$$
-300int_0^1x^4ln(1-x)dx = -300int_0^1(1-x)^4ln(x)dx = 300left(1 - frac{4}{4} + frac{6}{9}-frac{4}{16}+frac{1}{25}right) = 137
$$
So, in conclusion,
$$
sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^infty frac{300}{2^nk+5} =137
$$

share|cite|improve this answer

answered Mar 24 at 6:26

eyeballfrogeyeballfrog

7,212633

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks so much! It really helped!
  $endgroup$
  – Lee Laindingold
  Mar 24 at 8:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually,this question appeared in a secondary school maths competition so I doubt any calculus is involved..................but at least i found a way of solving it, thanks!It would also be helpful if anyone could provide a more secondary school solution haha
  $endgroup$
  – Lee Laindingold
  Mar 24 at 8:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LeeLaindingold Infinite series are also calculus, and I believe they are usually taught after elementary integrals.
  $endgroup$
  – eyeballfrog
  Mar 24 at 8:12
  

  
  
  
  

add a comment | 

1

$begingroup$

This becomes easier by converting to an integral to get that weird linear term out of the denominator. Jack gave the quick version in the comments, but here's the more detailed explanation.

First, introduce an integral to convert from a zeta-like series to a power series:
begin{multline}
sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^infty frac{300}{2^nk+5} = 300sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^inftyint_0^1 x^{2^nk+4}dx
\= 300int_0^1x^4left[sum_{n=0}^inftysum_{k=1}^infty (-1)^{k-1}frac{x^{2^nk}}{k}right] dx = 300int_0^1x^4left[sum_{n=0}^inftylnleft(1+x^{2^n}right)right] dx,
end{multline}
where we used the series expansion $ln(1+z) = sum_{k=0}^infty (-1)^{k-1}z^k/k$ to do the $k$ summation.

Next, evaluate the $n$ series using
$$
sum_{n=0}^infty ln(1+x^{2^n}) = lnleft[prod_{n=0}^infty left(1+x^{2^n}right)right] = lnleft[sum_{m=0}^infty x^mright] = lnleft(frac{1}{1-x}right) = -ln(1-x).
$$
The identity with the infinite product can be seen by considering expanding each exponent $m$ as sums of powers of $2$. So now we've got our sum in terms of a nice integral,
$$
sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^infty frac{300}{2^nk+5} = -300int_0^1x^4ln(1-x)dx
$$
Last step is to use the identity $int_0^1 x^nln(x)dx =-(n+1)^{-2}$ to get
$$
-300int_0^1x^4ln(1-x)dx = -300int_0^1(1-x)^4ln(x)dx = 300left(1 - frac{4}{4} + frac{6}{9}-frac{4}{16}+frac{1}{25}right) = 137
$$
So, in conclusion,
$$
sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^infty frac{300}{2^nk+5} =137
$$

share|cite|improve this answer

answered Mar 24 at 6:26

eyeballfrogeyeballfrog

7,212633

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks so much! It really helped!
  $endgroup$
  – Lee Laindingold
  Mar 24 at 8:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually,this question appeared in a secondary school maths competition so I doubt any calculus is involved..................but at least i found a way of solving it, thanks!It would also be helpful if anyone could provide a more secondary school solution haha
  $endgroup$
  – Lee Laindingold
  Mar 24 at 8:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LeeLaindingold Infinite series are also calculus, and I believe they are usually taught after elementary integrals.
  $endgroup$
  – eyeballfrog
  Mar 24 at 8:12
  

  
  
  
  

add a comment | 

$begingroup$

This becomes easier by converting to an integral to get that weird linear term out of the denominator. Jack gave the quick version in the comments, but here's the more detailed explanation.

First, introduce an integral to convert from a zeta-like series to a power series:
begin{multline}
sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^infty frac{300}{2^nk+5} = 300sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^inftyint_0^1 x^{2^nk+4}dx
\= 300int_0^1x^4left[sum_{n=0}^inftysum_{k=1}^infty (-1)^{k-1}frac{x^{2^nk}}{k}right] dx = 300int_0^1x^4left[sum_{n=0}^inftylnleft(1+x^{2^n}right)right] dx,
end{multline}
where we used the series expansion $ln(1+z) = sum_{k=0}^infty (-1)^{k-1}z^k/k$ to do the $k$ summation.

Next, evaluate the $n$ series using
$$
sum_{n=0}^infty ln(1+x^{2^n}) = lnleft[prod_{n=0}^infty left(1+x^{2^n}right)right] = lnleft[sum_{m=0}^infty x^mright] = lnleft(frac{1}{1-x}right) = -ln(1-x).
$$
The identity with the infinite product can be seen by considering expanding each exponent $m$ as sums of powers of $2$. So now we've got our sum in terms of a nice integral,
$$
sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^infty frac{300}{2^nk+5} = -300int_0^1x^4ln(1-x)dx
$$
Last step is to use the identity $int_0^1 x^nln(x)dx =-(n+1)^{-2}$ to get
$$
-300int_0^1x^4ln(1-x)dx = -300int_0^1(1-x)^4ln(x)dx = 300left(1 - frac{4}{4} + frac{6}{9}-frac{4}{16}+frac{1}{25}right) = 137
$$
So, in conclusion,
$$
sum_{k=1}^inftyfrac{(-1)^{k-1}}{k}sum_{n=0}^infty frac{300}{2^nk+5} =137
$$

share|cite|improve this answer

answered Mar 24 at 6:26

eyeballfrogeyeballfrog

7,212633

$endgroup$

share|cite|improve this answer

answered Mar 24 at 6:26

eyeballfrogeyeballfrog

7,212633

answered Mar 24 at 6:26

eyeballfrogeyeballfrog

7,212633

answered Mar 24 at 6:26

eyeballfrogeyeballfrog

7,212633