# Suppose $\lim\limits_{n \rightarrow \infty } n^2 a_n =1$, then $\sum\limits _{n=1} ^{\infty} a_n$ is convergent? Or is it divergent?

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.

• you argument is correct Commented Dec 1, 2018 at 14:52
• 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 Commented Dec 1, 2018 at 14:54
• @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.
– Did
Commented Dec 1, 2018 at 15:09
• Of course, it is bounded AND increasing, therefore it converges!
– user459879
Commented Dec 1, 2018 at 15:11
• @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...)
– Did
Commented Dec 1, 2018 at 15:31

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.

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.

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.

• It is increasing and bounded, therefore it converges!
– user459879
Commented Dec 1, 2018 at 15:10
• Of course, but equivalence gives a faster proof. Bondedness is implicit in the definition of equivalence. Commented Dec 1, 2018 at 15:16

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.

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