Is it true that $\sum^{\infty}_{n=1}a_n$ converges $\iff$ $\sum^{\infty}_{n=1}(1 + \frac{1}{n})a_n$ converges? I think that I can prove the $\implies$ implication: $$\sum^{\infty}_{n=1}(1 + \frac{1}{n})a_n = \sum^{\infty}_{n=1}(a_n+\frac{a_n}{n})$$ We know that $\sum^{\infty}_{n=1}\frac{a_n}{n}$ converges from Dirichlet's test so the sum of these two series is finite so the series above converges. I don't have any idea how to prove <= implication. Any hints?

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    $\begingroup$ do you know whether $a_n$ is positive? $\endgroup$ – alonkol Feb 3 '15 at 22:38
  • $\begingroup$ Are you sure that $\sum_n(a_n)+\sum_n(\frac{a_n}{n})=\sum_n(a_n+\frac{a_n}{n})$? $\endgroup$ – Steven Stadnicki Feb 3 '15 at 22:38
  • $\begingroup$ @alonkol $a_n$ can be any sequence $\endgroup$ – wisniak Feb 3 '15 at 22:45
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    $\begingroup$ @StevenStadnicki I've used arithmetic properties of limits $\endgroup$ – wisniak Feb 3 '15 at 22:46
  • $\begingroup$ What if $\sum a_n$ convergent, $\epsilon_n \to 0$ but not decreasing, does it imply $\sum (1+\epsilon_n) a_n$ convergent ? $\endgroup$ – Orest Bucicovschi Feb 4 '15 at 0:14

Dirichlet's test, as you know, says that if $f(n)$ is a decreasing positive function, and $\sum a_n$ converges (or even just boundedly oscillates), then $\sum f(n)a_n$ converges. Assuming that $\sum a_n$ converges, you applied this with $f(n)=1/n$ to show that $\sum a_n/n$ converges; your implication then follows from $\sum a_n + \sum a_n/n = \sum (1+1/n)a_n$.

Conversely, assume that $\sum (1+1/n)a_n$ converges. Define $f(n)=1/(n+1)$; then $\sum f(n)(1+1/n)a_n = \sum a_n/n$ converges by Dirichlet's test again. The converse implication then follows from $\sum (1+1/n)a_n - \sum a_n/n = \sum a_n$.

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    $\begingroup$ Perhaps you wanted $\sum_n f(n) (1+1/n) a_n = \sum_n a_n/n$? $\endgroup$ – dtldarek Feb 3 '15 at 23:01
  • $\begingroup$ Good catch - fixed! $\endgroup$ – Greg Martin Feb 3 '15 at 23:49

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