Discuss the convergence of the sequence $(a_{n})$, where $a_{n}:=1+\frac{1}{2^2}+\frac{1}{3^2}+ \dots +\frac{1}{n^2}$. How to show this

Question

Discuss the convergence of the sequence $(a_{n})$, where $a_{n}:=1+\frac{1}{2^2}+\frac{1}{3^2}+ \dots +\frac{1}{n^2}$?

Do you have any ideas? What tools do you have to analyse convergence of series? — Arthur, Oct 05 '16 at 06:54
I was trying to show that this sequence is Cauchy sequence. We can write this as \sum_{k=0}^{n}1/k^2 and use definition of Cauchy — user375201, Oct 05 '16 at 06:59
Using Cauchy will be a little bit of a circle. Since for all $\epsilon >0$ you must find a $M$ such that $\sum_{k=0}^n \frac{1}{k^2} - \sum_{k=0}^m\frac{1}{k^2} < \epsilon$ for all $n \ge m \ge M$. Taking $m=M$, and $n$ really big, you show $\sum_{k=M}^\infty \frac{1}{k^2} < \epsilon$, which is practically that same as showing $\sum_{k=1}^\infty \frac{1}{k^2} < \infty$, for which you want to use Cauchy, so your back at square 1. — Hetebrij, Oct 05 '16 at 07:22

score 1 · Answer 1 · 2016-10-05T08:00:20.973

1

$$\frac1{k^2}<\frac1{(k-1)k}=\frac1{k-1}-\frac1k$$ and by telescoping, the sum (but the term $k=1$) is bounded by $$1-\frac1n.$$

edited Oct 05 '16 at 08:00

answered Oct 05 '16 at 07:57

@Cm7F7Bb Thanks for your interest in this answer. The telescopic sum avoids the recourse to an integral. – Oct 05 '16 at 08:19

score 1 · Answer 2 · answered Oct 05 '16 at 08:45

1

Use cauchy condensation test, whic says $\Sigma a_{n}$ converges iff $\Sigma 2^{n}a_{2^{n}}$ converges. Thus $\Sigma \frac{1}{n^{2}}$ converges iff $\Sigma \frac{2^{n}}{2^{2n}}$ converges. ie, iff $\Sigma\frac{1}{2^{n}}$ converges. We know the latter series is a geometric series which converges to 2.

answered Oct 05 '16 at 08:45

Abdul Jaleel A P

182

($a_n$ positive decreasing) tks I didn't know this way for showing the convergence of Riemann integrals – reuns Oct 05 '16 at 08:54

score 0 · Answer 3 · answered Oct 05 '16 at 07:44

It's a p-series, and if p>1, it converges. No need to do anything more elaborate, the tags on this question are calculus and sequences/series, not analysis (not that the post below me is wrong, or anything, of course; and my apologies if the OP is doing this for a more advanced course. This explanation is perfect for calculus however).

score 0 · Answer 4 · answered Oct 05 '16 at 07:44

0

By approximating the sum by a definite integral from above, $$ \sum_{k=1}^n\frac1{k^2}=1+\sum_{k=2}^n\frac1{k^2}\int_{k-1}^k\mathrm dx\le1+\int_1^n\frac1{x^2}\mathrm dx=1+[-x^{-1}]_1^n=2-\frac1n. $$ The sequence $$ \sum_{k=1}^n\frac1{k^2} $$ is bounded. Hence, it converges.

answered Oct 05 '16 at 07:44

Cm7F7Bb

17,364

But we have the result that every convergence sequence is bounded but converse not trure. for e.g. take $(a_n)=(-1)^{n}$. It is bounded but not convergent. – user375201 Oct 06 '16 at 07:01
@user375201 Yes, you are right. However, a bounded monotone sequence does converge and the sequence $\sum_{k=1}^nk^{-2}$ is monotone. – Cm7F7Bb Oct 06 '16 at 07:08
Yes. now its fine. – user375201 Oct 06 '16 at 10:44

score -3 · Answer 5 · answered Oct 05 '16 at 07:09

-3

$$\lim_{n \to \infty} (a_{n}) = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k^{2}} = \zeta(2) = \frac{\pi ^{2}}{6}$$

Where $\zeta(s)$ is the Riemann zeta function.

answered Oct 05 '16 at 07:09

poweierstrass

2,388

This does not give the OP any clue or guidance on how to prove the convergence.. – Joshhh Oct 05 '16 at 07:15
It points the OP in the right direction by noting that it is a well known function and so decreases the time it would take to find more information about convergence. – poweierstrass Oct 05 '16 at 07:21
@poweierstrass: with this reasoning, just point to the "Basel problem". – Oct 05 '16 at 07:55
We have to show by definition of Cauchy sequence – user375201 Oct 05 '16 at 07:57

Discuss the convergence of the sequence $(a_{n})$, where $a_{n}:=1+\frac{1}{2^2}+\frac{1}{3^2}+ \dots +\frac{1}{n^2}$. How to show this

5 Answers5