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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?

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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.

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    $\begingroup$ It's probably some epsilon-delta extremist $\endgroup$
    – Snoop
    Apr 20 at 19:55
  • $\begingroup$ Thank you. But would my solution be valid? $\endgroup$
    – Veritas
    Apr 20 at 20:04
  • $\begingroup$ In en.wikipedia.org/wiki/Convergence_tests#Ratio_test you can see, that if limit equals $1$, then it is inconclusive. $\endgroup$
    – zkutch
    Apr 20 at 20:07
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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 $y_1=y_2=0$.

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  • $\begingroup$ Thank you. But would my solution be valid? $\endgroup$
    – Veritas
    Apr 20 at 20:04
  • $\begingroup$ It looks solid, but you should prove that $b_n=(n+1)(n-2)/((n-1)n)$ converges to 1. $\endgroup$
    – Snoop
    Apr 20 at 20:12
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Let $n>4;$

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

Absolutely convergent (comparison test).

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