Show that a Series Converges using Induction and MCT. The Next CEO of Stack OverflowProve that the series convergesProve that if $a_n^2$ converges ($a_n$ is monotone), thus $a_n$ converges and to what?Show that the series $sum_k=0 ^infty (-1)^k fracx^2k+12k+1$ converges for $|x|<1$ and that it converges to $arctan x$prove series converges for all $p>1$infinite series and proof of sum using induction.Prove that the subsequence of a convergent series with nonnegative terms convergesProve that $a_2n+1$ and $a_2n$ are convergent, and that $a_2n+1-a_2n to 0$ when $n to infty$.Using induction & subsequences to prove the Alternating Series TestHow to prove that a recursive sequence converges?Let $sum_k=1^infty a_n$ be convergent show that $sum_k=1^infty n(a_n-a_n+1)$ converges
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Show that a Series Converges using Induction and MCT.
The Next CEO of Stack OverflowProve that the series convergesProve that if $a_n^2$ converges ($a_n$ is monotone), thus $a_n$ converges and to what?Show that the series $sum_k=0 ^infty (-1)^k fracx^2k+12k+1$ converges for $|x|<1$ and that it converges to $arctan x$prove series converges for all $p>1$infinite series and proof of sum using induction.Prove that the subsequence of a convergent series with nonnegative terms convergesProve that $a_2n+1$ and $a_2n$ are convergent, and that $a_2n+1-a_2n to 0$ when $n to infty$.Using induction & subsequences to prove the Alternating Series TestHow to prove that a recursive sequence converges?Let $sum_k=1^infty a_n$ be convergent show that $sum_k=1^infty n(a_n-a_n+1)$ converges
$begingroup$
Question: Show that the series $sum_n=0^infty frac(-1)^nn$ converges as follows. Let Sn denote the nth partial sums. Show that S2n−1 is a bounded monotone sequence and therefore convergent. Then show that Sn actually converges by noting that the difference between S2n and S2n−1 is small.
Let an = $frac(-1)^nn$, Sn = $sum_n=0^infty$ an
a1 = -1 --------------------- a2 = 1/2
a3 = -1/3 ------------------ a4 = 1/4
a5 = -1/5 ------------------ a6 = 1/6
. .
. .
. .
S2n-1 = $sum_n=0^infty$ a2n-1
S1 = -1
S2 = -1/3
S3 = -1/5
Claim: S2n-1 is monotone increasing
S1 < S2
S2 < S3
Prove using induction:
Let P(n): S2n-1 < S2n
P(1): S1 < S2 : -1 < -1/3 (True)
P(2): S3 < S4 : -1/5 < -1/7 (True)
This is what I have got so far, can anyone help me out because I'm lost, the induction is confusing for me, thank you very much.
sequences-and-series convergence induction
asked Mar 18 at 12:06
Benson TanBenson Tan
1
$endgroup$
add a comment |
$begingroup$
Question: Show that the series $sum_n=0^infty frac(-1)^nn$ converges as follows. Let Sn denote the nth partial sums. Show that S2n−1 is a bounded monotone sequence and therefore convergent. Then show that Sn actually converges by noting that the difference between S2n and S2n−1 is small.
Let an = $frac(-1)^nn$, Sn = $sum_n=0^infty$ an
a1 = -1 --------------------- a2 = 1/2
a3 = -1/3 ------------------ a4 = 1/4
a5 = -1/5 ------------------ a6 = 1/6
. .
. .
. .
S2n-1 = $sum_n=0^infty$ a2n-1
S1 = -1
S2 = -1/3
S3 = -1/5
Claim: S2n-1 is monotone increasing
S1 < S2
S2 < S3
Prove using induction:
Let P(n): S2n-1 < S2n
P(1): S1 < S2 : -1 < -1/3 (True)
P(2): S3 < S4 : -1/5 < -1/7 (True)
This is what I have got so far, can anyone help me out because I'm lost, the induction is confusing for me, thank you very much.
sequences-and-series convergence induction
asked Mar 18 at 12:06
Benson TanBenson Tan
1
$endgroup$
$begingroup$
Hi. Can you check and confirm that you understand what a partial sum is?
$endgroup$
– user66081
Mar 18 at 12:54
add a comment |
$begingroup$
Question: Show that the series $sum_n=0^infty frac(-1)^nn$ converges as follows. Let Sn denote the nth partial sums. Show that S2n−1 is a bounded monotone sequence and therefore convergent. Then show that Sn actually converges by noting that the difference between S2n and S2n−1 is small.
Let an = $frac(-1)^nn$, Sn = $sum_n=0^infty$ an
a1 = -1 --------------------- a2 = 1/2
a3 = -1/3 ------------------ a4 = 1/4
a5 = -1/5 ------------------ a6 = 1/6
. .
. .
. .
S2n-1 = $sum_n=0^infty$ a2n-1
S1 = -1
S2 = -1/3
S3 = -1/5
Claim: S2n-1 is monotone increasing
S1 < S2
S2 < S3
Prove using induction:
Let P(n): S2n-1 < S2n
P(1): S1 < S2 : -1 < -1/3 (True)
P(2): S3 < S4 : -1/5 < -1/7 (True)
This is what I have got so far, can anyone help me out because I'm lost, the induction is confusing for me, thank you very much.
sequences-and-series convergence induction
asked Mar 18 at 12:06
Benson TanBenson Tan
1
$endgroup$
Question: Show that the series $sum_n=0^infty frac(-1)^nn$ converges as follows. Let Sn denote the nth partial sums. Show that S2n−1 is a bounded monotone sequence and therefore convergent. Then show that Sn actually converges by noting that the difference between S2n and S2n−1 is small.
Let an = $frac(-1)^nn$, Sn = $sum_n=0^infty$ an
a1 = -1 --------------------- a2 = 1/2
a3 = -1/3 ------------------ a4 = 1/4
a5 = -1/5 ------------------ a6 = 1/6
. .
. .
. .
S2n-1 = $sum_n=0^infty$ a2n-1
S1 = -1
S2 = -1/3
S3 = -1/5
Claim: S2n-1 is monotone increasing
S1 < S2
S2 < S3
Prove using induction:
Let P(n): S2n-1 < S2n
P(1): S1 < S2 : -1 < -1/3 (True)
P(2): S3 < S4 : -1/5 < -1/7 (True)
This is what I have got so far, can anyone help me out because I'm lost, the induction is confusing for me, thank you very much.
sequences-and-series convergence induction
sequences-and-series convergence induction
asked Mar 18 at 12:06
Benson TanBenson Tan
1
asked Mar 18 at 12:06
Benson TanBenson Tan
1
asked Mar 18 at 12:06
Benson TanBenson Tan
1
asked Mar 18 at 12:06
Benson TanBenson Tan
1
asked Mar 18 at 12:06
Benson TanBenson Tan
1
1
$begingroup$
Hi. Can you check and confirm that you understand what a partial sum is?
$endgroup$
– user66081
Mar 18 at 12:54
add a comment |
$begingroup$
Hi. Can you check and confirm that you understand what a partial sum is?
$endgroup$
– user66081
Mar 18 at 12:54
$begingroup$
Hi. Can you check and confirm that you understand what a partial sum is?
$endgroup$
– user66081
Mar 18 at 12:54
$begingroup$
Hi. Can you check and confirm that you understand what a partial sum is?
$endgroup$
– user66081
Mar 18 at 12:54
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The series converges according to the Leibniz's alternating series test. https://en.wikipedia.org/wiki/Alternating_series_test .
The OP's method also works as follows:
beginalign
b_n&=S_2n=sum_k=1^nleft(-frac12k-1+frac12kright)=b_n-1-frac1(2n-1)2n<b_n-1\
b_n&=sum_k=1^nfrac-1(2k-1)2k>sum_k=1^nfrac-1(2k-1)^2>-1-sum_k=2^nfrac1(2k-2)^2=-1-frac14sum_k=1^n-1frac1k^2\
&>-1-fracpi^224\\
textSo b_n&text is decreasing but lower bounded. Similarly,\\
c_n&=S_2n+1=-1+sum_k=1^nleft(frac12k-frac12k+1right)=c_n-1+frac12n(2n+1)>b_n-1\
c_n&=-1+sum_k=1^nfrac12k(2k+1)<-1+sum_k=1^nfrac1(2k)^2=-1+frac14sum_k=1^nfrac1k^2<-1+fracpi^224\\
textSo c_n&text is increasing but upper bounded.\\
endalign
Therefore, both $S_2n$ and $S_2n+1$ converges. And as both of those converge,
beginalign
0&=lim_ntoinftyleft(S_2n-S_2n+1right)=lim_ntoinftyS_2n-lim_ntoinftyS_2n+1\
therefore&lim_ntoinftyS_2n=lim_ntoinftyS_2n+1
endalign
So $S_n$ converges.
answered Mar 18 at 18:41
Kay K.Kay K.
6,9401337
$endgroup$
$begingroup$
Hi, thanks for answering the question! But I don't know how you got your S2n? because what I got for my induction so far is, Assume P(K) : S2n-1 = $sum_k=1^n frac(-12k-1$ is true. Consider P(K+1) : S2n = $sum_k=1^n+1 frac(-12k-1$ = S2n = $[sum_k=1^n frac(-12k-1][frac(-12n]$
$endgroup$
– Benson Tan
Mar 19 at 7:00
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
The series converges according to the Leibniz's alternating series test. https://en.wikipedia.org/wiki/Alternating_series_test .
The OP's method also works as follows:
beginalign
b_n&=S_2n=sum_k=1^nleft(-frac12k-1+frac12kright)=b_n-1-frac1(2n-1)2n<b_n-1\
b_n&=sum_k=1^nfrac-1(2k-1)2k>sum_k=1^nfrac-1(2k-1)^2>-1-sum_k=2^nfrac1(2k-2)^2=-1-frac14sum_k=1^n-1frac1k^2\
&>-1-fracpi^224\\
textSo b_n&text is decreasing but lower bounded. Similarly,\\
c_n&=S_2n+1=-1+sum_k=1^nleft(frac12k-frac12k+1right)=c_n-1+frac12n(2n+1)>b_n-1\
c_n&=-1+sum_k=1^nfrac12k(2k+1)<-1+sum_k=1^nfrac1(2k)^2=-1+frac14sum_k=1^nfrac1k^2<-1+fracpi^224\\
textSo c_n&text is increasing but upper bounded.\\
endalign
Therefore, both $S_2n$ and $S_2n+1$ converges. And as both of those converge,
beginalign
0&=lim_ntoinftyleft(S_2n-S_2n+1right)=lim_ntoinftyS_2n-lim_ntoinftyS_2n+1\
therefore&lim_ntoinftyS_2n=lim_ntoinftyS_2n+1
endalign
So $S_n$ converges.
answered Mar 18 at 18:41
Kay K.Kay K.
6,9401337
$endgroup$
$begingroup$
Hi, thanks for answering the question! But I don't know how you got your S2n? because what I got for my induction so far is, Assume P(K) : S2n-1 = $sum_k=1^n frac(-12k-1$ is true. Consider P(K+1) : S2n = $sum_k=1^n+1 frac(-12k-1$ = S2n = $[sum_k=1^n frac(-12k-1][frac(-12n]$
$endgroup$
– Benson Tan
Mar 19 at 7:00
add a comment |
$begingroup$
The series converges according to the Leibniz's alternating series test. https://en.wikipedia.org/wiki/Alternating_series_test .
The OP's method also works as follows:
beginalign
b_n&=S_2n=sum_k=1^nleft(-frac12k-1+frac12kright)=b_n-1-frac1(2n-1)2n<b_n-1\
b_n&=sum_k=1^nfrac-1(2k-1)2k>sum_k=1^nfrac-1(2k-1)^2>-1-sum_k=2^nfrac1(2k-2)^2=-1-frac14sum_k=1^n-1frac1k^2\
&>-1-fracpi^224\\
textSo b_n&text is decreasing but lower bounded. Similarly,\\
c_n&=S_2n+1=-1+sum_k=1^nleft(frac12k-frac12k+1right)=c_n-1+frac12n(2n+1)>b_n-1\
c_n&=-1+sum_k=1^nfrac12k(2k+1)<-1+sum_k=1^nfrac1(2k)^2=-1+frac14sum_k=1^nfrac1k^2<-1+fracpi^224\\
textSo c_n&text is increasing but upper bounded.\\
endalign
Therefore, both $S_2n$ and $S_2n+1$ converges. And as both of those converge,
beginalign
0&=lim_ntoinftyleft(S_2n-S_2n+1right)=lim_ntoinftyS_2n-lim_ntoinftyS_2n+1\
therefore&lim_ntoinftyS_2n=lim_ntoinftyS_2n+1
endalign
So $S_n$ converges.
answered Mar 18 at 18:41
Kay K.Kay K.
6,9401337
$endgroup$
$begingroup$
Hi, thanks for answering the question! But I don't know how you got your S2n? because what I got for my induction so far is, Assume P(K) : S2n-1 = $sum_k=1^n frac(-12k-1$ is true. Consider P(K+1) : S2n = $sum_k=1^n+1 frac(-12k-1$ = S2n = $[sum_k=1^n frac(-12k-1][frac(-12n]$
$endgroup$
– Benson Tan
Mar 19 at 7:00
add a comment |
$begingroup$
The series converges according to the Leibniz's alternating series test. https://en.wikipedia.org/wiki/Alternating_series_test .
The OP's method also works as follows:
beginalign
b_n&=S_2n=sum_k=1^nleft(-frac12k-1+frac12kright)=b_n-1-frac1(2n-1)2n<b_n-1\
b_n&=sum_k=1^nfrac-1(2k-1)2k>sum_k=1^nfrac-1(2k-1)^2>-1-sum_k=2^nfrac1(2k-2)^2=-1-frac14sum_k=1^n-1frac1k^2\
&>-1-fracpi^224\\
textSo b_n&text is decreasing but lower bounded. Similarly,\\
c_n&=S_2n+1=-1+sum_k=1^nleft(frac12k-frac12k+1right)=c_n-1+frac12n(2n+1)>b_n-1\
c_n&=-1+sum_k=1^nfrac12k(2k+1)<-1+sum_k=1^nfrac1(2k)^2=-1+frac14sum_k=1^nfrac1k^2<-1+fracpi^224\\
textSo c_n&text is increasing but upper bounded.\\
endalign
Therefore, both $S_2n$ and $S_2n+1$ converges. And as both of those converge,
beginalign
0&=lim_ntoinftyleft(S_2n-S_2n+1right)=lim_ntoinftyS_2n-lim_ntoinftyS_2n+1\
therefore&lim_ntoinftyS_2n=lim_ntoinftyS_2n+1
endalign
So $S_n$ converges.
answered Mar 18 at 18:41
Kay K.Kay K.
6,9401337
$endgroup$
The series converges according to the Leibniz's alternating series test. https://en.wikipedia.org/wiki/Alternating_series_test .
The OP's method also works as follows:
beginalign
b_n&=S_2n=sum_k=1^nleft(-frac12k-1+frac12kright)=b_n-1-frac1(2n-1)2n<b_n-1\
b_n&=sum_k=1^nfrac-1(2k-1)2k>sum_k=1^nfrac-1(2k-1)^2>-1-sum_k=2^nfrac1(2k-2)^2=-1-frac14sum_k=1^n-1frac1k^2\
&>-1-fracpi^224\\
textSo b_n&text is decreasing but lower bounded. Similarly,\\
c_n&=S_2n+1=-1+sum_k=1^nleft(frac12k-frac12k+1right)=c_n-1+frac12n(2n+1)>b_n-1\
c_n&=-1+sum_k=1^nfrac12k(2k+1)<-1+sum_k=1^nfrac1(2k)^2=-1+frac14sum_k=1^nfrac1k^2<-1+fracpi^224\\
textSo c_n&text is increasing but upper bounded.\\
endalign
Therefore, both $S_2n$ and $S_2n+1$ converges. And as both of those converge,
beginalign
0&=lim_ntoinftyleft(S_2n-S_2n+1right)=lim_ntoinftyS_2n-lim_ntoinftyS_2n+1\
therefore&lim_ntoinftyS_2n=lim_ntoinftyS_2n+1
endalign
So $S_n$ converges.
answered Mar 18 at 18:41
Kay K.Kay K.
6,9401337
answered Mar 18 at 18:41
Kay K.Kay K.
6,9401337
answered Mar 18 at 18:41
Kay K.Kay K.
6,9401337
answered Mar 18 at 18:41
Kay K.Kay K.
6,9401337
6,9401337
$begingroup$
Hi, thanks for answering the question! But I don't know how you got your S2n? because what I got for my induction so far is, Assume P(K) : S2n-1 = $sum_k=1^n frac(-12k-1$ is true. Consider P(K+1) : S2n = $sum_k=1^n+1 frac(-12k-1$ = S2n = $[sum_k=1^n frac(-12k-1][frac(-12n]$
$endgroup$
– Benson Tan
Mar 19 at 7:00
add a comment |
$begingroup$
Hi, thanks for answering the question! But I don't know how you got your S2n? because what I got for my induction so far is, Assume P(K) : S2n-1 = $sum_k=1^n frac(-12k-1$ is true. Consider P(K+1) : S2n = $sum_k=1^n+1 frac(-12k-1$ = S2n = $[sum_k=1^n frac(-12k-1][frac(-12n]$
$endgroup$
– Benson Tan
Mar 19 at 7:00
$begingroup$
Hi, thanks for answering the question! But I don't know how you got your S2n? because what I got for my induction so far is, Assume P(K) : S2n-1 = $sum_k=1^n frac(-12k-1$ is true. Consider P(K+1) : S2n = $sum_k=1^n+1 frac(-12k-1$ = S2n = $[sum_k=1^n frac(-12k-1][frac(-12n]$
$endgroup$
– Benson Tan
Mar 19 at 7:00
$begingroup$
Hi, thanks for answering the question! But I don't know how you got your S2n? because what I got for my induction so far is, Assume P(K) : S2n-1 = $sum_k=1^n frac(-12k-1$ is true. Consider P(K+1) : S2n = $sum_k=1^n+1 frac(-12k-1$ = S2n = $[sum_k=1^n frac(-12k-1][frac(-12n]$
$endgroup$
– Benson Tan
Mar 19 at 7:00
add a comment |
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$begingroup$
Hi. Can you check and confirm that you understand what a partial sum is?
$endgroup$
– user66081
Mar 18 at 12:54