Finding a sequences upper and lower bounds The 2019 Stack Overflow Developer Survey Results...
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Finding a sequences upper and lower bounds
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove ${b_n} = sup { {a_k}:k geqslant n} $for $n geqslant 1$. Prove that $({b_n})$ converges.Having trouble understanding Series and SequencesProve the convergence of a series.Find $limlimits_{ntoinfty}left(frac{a_1}{a_2}+frac{a_2}{a_3}+frac{a_3}{a_4}+…+frac{a_n}{a_1}right)$Prove that if $a_1 + a_2 + ldots$ converges then $a_1+2a_2+4a_4+8 a_8+ldots$ converges and $lim na_n=0$Show that $a_n^{1/n}$ converges as $ntoinfty$First term of a series with two zeros and a constant second differenceProving the Alternate Series Test with Monotone ConvergenceUsing induction & subsequences to prove the Alternating Series TestLimit involving an arithmetic progression
$begingroup$
I have a series defined as $a_n=frac{1}{(n+1)}+frac{1}{(n+2)}+...+ frac{1}{(2n)}$ for $ngeq1$.
How can I show that $a_nleqfrac{n}{(n+1)}$ and that it is also bounded?
So far I have tried to work out the first few terms like $a_1=frac{1}{2}$ , $a_2=frac{7}{12}$, $a_3=frac{37}{60}$, $a_4=frac{533}{840}$ so I can see it is increasing but cannot see to show the above algebraically!
I know there is a question similar but I am not working out whether it converges or not
real-analysis calculus sequences-and-series
asked Mar 22 at 22:17
James odareJames odare
989
$endgroup$
add a comment |
$begingroup$
I have a series defined as $a_n=frac{1}{(n+1)}+frac{1}{(n+2)}+...+ frac{1}{(2n)}$ for $ngeq1$.
How can I show that $a_nleqfrac{n}{(n+1)}$ and that it is also bounded?
So far I have tried to work out the first few terms like $a_1=frac{1}{2}$ , $a_2=frac{7}{12}$, $a_3=frac{37}{60}$, $a_4=frac{533}{840}$ so I can see it is increasing but cannot see to show the above algebraically!
I know there is a question similar but I am not working out whether it converges or not
real-analysis calculus sequences-and-series
asked Mar 22 at 22:17
James odareJames odare
989
$endgroup$
3
$begingroup$
You have a sum of $n$ terms, and each term is $le frac{1}{n+1}$, so ...
$endgroup$
– Martin R
Mar 22 at 22:20
$begingroup$
I see, but it there a way of working it out rather than using intuition
$endgroup$
– James odare
Mar 22 at 22:21
2
$begingroup$
That is working it out, that isn't intuition.
$endgroup$
– kingW3
Mar 22 at 22:32
add a comment |
$begingroup$
I have a series defined as $a_n=frac{1}{(n+1)}+frac{1}{(n+2)}+...+ frac{1}{(2n)}$ for $ngeq1$.
How can I show that $a_nleqfrac{n}{(n+1)}$ and that it is also bounded?
So far I have tried to work out the first few terms like $a_1=frac{1}{2}$ , $a_2=frac{7}{12}$, $a_3=frac{37}{60}$, $a_4=frac{533}{840}$ so I can see it is increasing but cannot see to show the above algebraically!
I know there is a question similar but I am not working out whether it converges or not
real-analysis calculus sequences-and-series
asked Mar 22 at 22:17
James odareJames odare
989
$endgroup$
I have a series defined as $a_n=frac{1}{(n+1)}+frac{1}{(n+2)}+...+ frac{1}{(2n)}$ for $ngeq1$.
How can I show that $a_nleqfrac{n}{(n+1)}$ and that it is also bounded?
So far I have tried to work out the first few terms like $a_1=frac{1}{2}$ , $a_2=frac{7}{12}$, $a_3=frac{37}{60}$, $a_4=frac{533}{840}$ so I can see it is increasing but cannot see to show the above algebraically!
I know there is a question similar but I am not working out whether it converges or not
real-analysis calculus sequences-and-series
real-analysis calculus sequences-and-series
asked Mar 22 at 22:17
James odareJames odare
989
asked Mar 22 at 22:17
James odareJames odare
989
asked Mar 22 at 22:17
James odareJames odare
989
asked Mar 22 at 22:17
James odareJames odare
989
asked Mar 22 at 22:17
James odareJames odare
989
989
3
$begingroup$
You have a sum of $n$ terms, and each term is $le frac{1}{n+1}$, so ...
$endgroup$
– Martin R
Mar 22 at 22:20
$begingroup$
I see, but it there a way of working it out rather than using intuition
$endgroup$
– James odare
Mar 22 at 22:21
2
$begingroup$
That is working it out, that isn't intuition.
$endgroup$
– kingW3
Mar 22 at 22:32
add a comment |
3
$begingroup$
You have a sum of $n$ terms, and each term is $le frac{1}{n+1}$, so ...
$endgroup$
– Martin R
Mar 22 at 22:20
$begingroup$
I see, but it there a way of working it out rather than using intuition
$endgroup$
– James odare
Mar 22 at 22:21
2
$begingroup$
That is working it out, that isn't intuition.
$endgroup$
– kingW3
Mar 22 at 22:32
3
3
$begingroup$
You have a sum of $n$ terms, and each term is $le frac{1}{n+1}$, so ...
$endgroup$
– Martin R
Mar 22 at 22:20
$begingroup$
You have a sum of $n$ terms, and each term is $le frac{1}{n+1}$, so ...
$endgroup$
– Martin R
Mar 22 at 22:20
$begingroup$
I see, but it there a way of working it out rather than using intuition
$endgroup$
– James odare
Mar 22 at 22:21
$begingroup$
I see, but it there a way of working it out rather than using intuition
$endgroup$
– James odare
Mar 22 at 22:21
2
2
$begingroup$
That is working it out, that isn't intuition.
$endgroup$
– kingW3
Mar 22 at 22:32
$begingroup$
That is working it out, that isn't intuition.
$endgroup$
– kingW3
Mar 22 at 22:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The sequence is increasing because, for each $n>1$,begin{align}a_n-a_{n-1}&=frac1{2n}+frac1{2n-1}-frac1n\&=frac1{2n(2n-1)}\&>0.end{align}And it is bounded because, for each natural $n$,begin{align}0&<a_n\&=frac1{n+1}+frac1{n+2}+cdots+frac1{2n}\&leqslantoverbrace{frac1{n+1}+cdots+frac1{n+1}}^{ntext{ times}}\&=frac n{n+1}\&<1.end{align}
answered Mar 22 at 22:29
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
add a comment |
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$begingroup$
The sequence is increasing because, for each $n>1$,begin{align}a_n-a_{n-1}&=frac1{2n}+frac1{2n-1}-frac1n\&=frac1{2n(2n-1)}\&>0.end{align}And it is bounded because, for each natural $n$,begin{align}0&<a_n\&=frac1{n+1}+frac1{n+2}+cdots+frac1{2n}\&leqslantoverbrace{frac1{n+1}+cdots+frac1{n+1}}^{ntext{ times}}\&=frac n{n+1}\&<1.end{align}
answered Mar 22 at 22:29
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
add a comment |
$begingroup$
The sequence is increasing because, for each $n>1$,begin{align}a_n-a_{n-1}&=frac1{2n}+frac1{2n-1}-frac1n\&=frac1{2n(2n-1)}\&>0.end{align}And it is bounded because, for each natural $n$,begin{align}0&<a_n\&=frac1{n+1}+frac1{n+2}+cdots+frac1{2n}\&leqslantoverbrace{frac1{n+1}+cdots+frac1{n+1}}^{ntext{ times}}\&=frac n{n+1}\&<1.end{align}
answered Mar 22 at 22:29
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
add a comment |
$begingroup$
The sequence is increasing because, for each $n>1$,begin{align}a_n-a_{n-1}&=frac1{2n}+frac1{2n-1}-frac1n\&=frac1{2n(2n-1)}\&>0.end{align}And it is bounded because, for each natural $n$,begin{align}0&<a_n\&=frac1{n+1}+frac1{n+2}+cdots+frac1{2n}\&leqslantoverbrace{frac1{n+1}+cdots+frac1{n+1}}^{ntext{ times}}\&=frac n{n+1}\&<1.end{align}
answered Mar 22 at 22:29
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
The sequence is increasing because, for each $n>1$,begin{align}a_n-a_{n-1}&=frac1{2n}+frac1{2n-1}-frac1n\&=frac1{2n(2n-1)}\&>0.end{align}And it is bounded because, for each natural $n$,begin{align}0&<a_n\&=frac1{n+1}+frac1{n+2}+cdots+frac1{2n}\&leqslantoverbrace{frac1{n+1}+cdots+frac1{n+1}}^{ntext{ times}}\&=frac n{n+1}\&<1.end{align}
answered Mar 22 at 22:29
José Carlos SantosJosé Carlos Santos
174k23134243
answered Mar 22 at 22:29
José Carlos SantosJosé Carlos Santos
174k23134243
answered Mar 22 at 22:29
José Carlos SantosJosé Carlos Santos
174k23134243
answered Mar 22 at 22:29
José Carlos SantosJosé Carlos Santos
174k23134243
174k23134243
add a comment |
add a comment |
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$begingroup$
You have a sum of $n$ terms, and each term is $le frac{1}{n+1}$, so ...
$endgroup$
– Martin R
Mar 22 at 22:20
$begingroup$
I see, but it there a way of working it out rather than using intuition
$endgroup$
– James odare
Mar 22 at 22:21
2
$begingroup$
That is working it out, that isn't intuition.
$endgroup$
– kingW3
Mar 22 at 22:32