Suppose a series $a_n$ is greater than 0 for all positive integer n, and that $\sum \frac {a_n} n$ converges, then does the following also converge? I was wondering if the following is true.
Suppose a series $a_n$ is greater than 0 for all positive integer n, and that $\sum \frac {a_n}n$ converges, then is $\displaystyle \lim_{m\to \infty}\sum_{n= 1}^m {a_n \over m+n} = 0$?
It seems to be true because if $\sum {a_n\over n}$ converges, then that means that ${a_n\over n }\to 0$ for $n\to \infty$. This means, neglecting $n$, ${a_n \over m+n}$ will also tend to 0, and thus the summation would be equal to 0, but I don't know if this is true.
 A: Let $\varepsilon$ be a strictly positive real.
$\sum\dfrac{a_n}{n}$ converges, so: $\displaystyle \ \ \exists N \in \mathbb N^{\star} \ , \ \sum_{n=N}^{+\infty} \dfrac{a_n}{n} \leqslant \dfrac{\varepsilon}{2}  $
Then:  $\ \ \displaystyle \forall m \geqslant N \ , \ \ \sum_{n=1}^{m} \dfrac{a_n}{m+n} \leqslant  \sum_{n=1}^{N-1} \dfrac{a_n}{m} + \sum_{n=N}^m \dfrac{a_n}{n} \leqslant \dfrac{\varepsilon}{2}+\dfrac{1}{m} \sum_{n=1}^{N-1} a_n $
Now: $\\ \displaystyle \exists M\geqslant N \ , \ \forall m\geqslant M , \dfrac{1}{m} \sum_{n=1}^{N-1} a_n \leqslant \dfrac{\varepsilon}{2}$
We can conclude that:
$$ \forall \varepsilon > 0 \ , \ \exists M \in \mathbb N \ , \ \forall m\geqslant M \ ,\ 0\leqslant \sum_{n=1}^m\dfrac{a_n}{n+m} \leqslant \varepsilon    $$
And  $\ \ \displaystyle \lim_{m\rightarrow +\infty} \sum_{n=1}^m\dfrac{a_n}{n+m} = 0$
A: let $v_n = \frac{a_n}{m+n}$, and $u_n = \frac{a_n}{n}$. Now observe that $\frac{a_n}{m+n} \le \frac{a_n}{2n}$, so by comparison test, we can say that as each term of $v_n$ is smaller or equal to that of $\frac{u_n}{2}$, and as $u_n$ converges, so $v_n$ also converges as $m \to \infty$.
I think this is it.
:)
