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Suppose $\lim_{n \rightarrow \infty } n^2 a_n =1$ then $ \sum _{n=1} ^{\infty} a_n$ is convergent or divergent?

We use the definition of the limit, then for $\varepsilon =1$ there must exist an $n_0\geq n$ such that $$ |n^2 a_n -1|<1$$ so we can say that: $$ 0 < n^2 a_n < 2 $$ We divide by: $$ 0 < a_n < \frac{2}{n^2} $$ We now know if we take the infinite sum we get: $$ 0 < \sum _{n=1} ^{\infty} a_n < \sum _{n=1} ^{\infty} \frac{2}{n^2} $$ We know that the right side converges, so the sum is bounded, but how would I prove that is converges? Or can we find a counterexample that diverges, bur stays within these bounds.

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    $\begingroup$ you argument is correct $\endgroup$
    – Guy Fsone
    Commented Dec 1, 2018 at 14:52
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    $\begingroup$ note that $a_n\ge 0$, hence the series converges, that is, the sequence defined by $s_n:=\sum_{k=0}^na_k$ is increasing and bounded, now apply Bolzano-Weierstrass theorem $\endgroup$
    – Masacroso
    Commented Dec 1, 2018 at 14:54
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    $\begingroup$ @Masacroso The sign of the terms $a_n$ is offtopic and not needed to conclude: if $|a_n|\leqslant c/n^2$ then $\sum a_n$ converges. $\endgroup$
    – Did
    Commented Dec 1, 2018 at 15:09
  • $\begingroup$ Of course, it is bounded AND increasing, therefore it converges! $\endgroup$
    – user459879
    Commented Dec 1, 2018 at 15:11
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    $\begingroup$ @Masacroso Why not give to the OP the proper tools they are lacking, from the onset, rather than letting them get only a fugitive glimpse of the picture? Anyway, as regards the case $b_n=(-1)^n$, your comment is misleading since then, the series $\sum b_n$ diverges hence one should not write things like $\sum\limits_{n=0}^\infty b_n$. (Actually, rereading your comment, I wonder if you are aware of the difference of nature between the series $\sum b_n$, which always exists, and the number $\sum\limits_{n=0}^\infty b_n$, which may or may not exist...) $\endgroup$
    – Did
    Commented Dec 1, 2018 at 15:31

4 Answers 4

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Option.

$\lim_{n \rightarrow \infty}n^2a_n=1.$

This Implies for $n \in \mathbb{Z^+}$ :

$n^2|a_n| \lt M$, positive real number.

$|a_n| \lt M/n^2.$

By comparison test $\sum |a_n|$ is convergent,

hence $\sum a_n$ is convergent.

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A series converges iff its remainder term $\sum\limits_N^\infty a_n \rightarrow 0$ as $N\rightarrow \infty$. Your argument shows that you can, for large enough $N$, bound $|\sum\limits_N^\infty a_n|$ by $\sum\limits_N^\infty \frac{1}{n^2}$. The latter term approaches zero as $N\rightarrow \infty$, since $\sum \frac{1}{n^2}$ converges.

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The hypothesis means first that, if $n$ is large enough, $a_n>0$, and also that $a_n$ is asymptotically equivalent to $\dfrac1{n^2}$.

Now two series with (eventually positive) equivalent terms both converge or both diverge.

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  • $\begingroup$ It is increasing and bounded, therefore it converges! $\endgroup$
    – user459879
    Commented Dec 1, 2018 at 15:10
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    $\begingroup$ Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence. $\endgroup$
    – Bernard
    Commented Dec 1, 2018 at 15:16
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Your argument is correct, more simply we can refer directly to limit comparison test with $$\sum \frac1{n^2}$$

indeed

$$\frac{a_n}{ \frac1{n^2}}=n^2a_n\to 1$$

therefore $\sum a_n$ converges.

It is important to recognize that since, more in general, when we solve a problem we don't need everytime to prove all the results that we have already proved in a general way.

Therefore if you are not requested to use esplicitely the $\epsilon-\delta$ definition a solution by limit comparison test is fine.

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    $\begingroup$ The test only applies to positive series, so a small step is needed before applying it. $\endgroup$ Commented Dec 1, 2018 at 15:10
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    $\begingroup$ @AndresE.Caicedo yes of course we can assume $a_n$ strictly positive for n sufficiently large. $\endgroup$
    – user
    Commented Dec 1, 2018 at 15:20

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