# Prove that this series is absolutely convergent.

For this question, I want to be able to either prove or disprove that

If $$a_n \not= 0 \forall n \geq 2$$ and $$\sum_{n=2}^{\infty}a_n$$ is absolutely convergent, then $$\sum_{n=2}^{\infty}\frac{na_n}{n-2}$$ is absolutely convergent.

I have not been able to find a counterexample to this, so I am assuming that this is true. In an attempt to prove this, I tried to use the Ratio Test, because the ratio test is what usually is used to determine absolute convergence. So we know that $$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|=L$$, and $$0 \leq L < 1$$. Applying the same logic, If I can show using the ratio test that $$\sum_{n=2}^{\infty}\frac{na_n}{n-2}$$ also has a limit of L, using the ratio test, then I'm done. It seems like that this method does work! I used the ratio test and got the following:

$$\lim_{n \to \infty}|\frac{(n+1)(n-2)}{(n-1)(n)} \frac{a_{n+1}}{a_n}|$$, since the fraction with the n's converges to one, the limit is still the same. So this series also absolutely converges. Does this constitute as a proof?

Hint: we have Limit comparison test $$\left|\frac{na_n}{n-2}\right| \sim |a_n|, n \to \infty$$

p.s. for someone who down votes answers based on asymptotic solutions: now it's good time to reveal all against arguments, if any.

• It's probably some epsilon-delta extremist Apr 20 at 19:55
• Thank you. But would my solution be valid? Apr 20 at 20:04
• In en.wikipedia.org/wiki/Convergence_tests#Ratio_test you can see, that if limit equals $1$, then it is inconclusive. Apr 20 at 20:07

Abel's test for convergence states that if $$\sum_{n=1}^\infty x_n$$ converges and $$(y_n)_{n \in \mathbb{N}}$$ is such that $$y_1\geq y_2\geq ... \geq 0$$ then the series $$\sum_{n=1}^\infty x_n y_n$$ converges.

In your example, the convergent series is $$\sum_{n=1}^\infty|a_n|$$ with $$a_1=a_2=0$$ and $$y_n=|y_n|=n/(n-2),\forall n > 2$$ so just set $$y_1=y_2=0$$.

• Thank you. But would my solution be valid? Apr 20 at 20:04
• It looks solid, but you should prove that $b_n=(n+1)(n-2)/((n-1)n)$ converges to 1. Apr 20 at 20:12

Let $$n>4;$$

$$|\dfrac{na_n}{n-2}| <| \dfrac{na_n}{n-n/2}| =2|a_n|.$$

Absolutely convergent (comparison test).