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?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.