Convergence of $\lim_{n \to \infty}\sum_{i=1}^{n}\frac{a_i}{i+n}$ If $\sum_{n=1}^{\infty}\frac{a_n}{n}=a$ converges to a finite number with $a_n \geq 0$ for all $n \geq 1$ does this also imply that $$\lim_{n \to \infty}\sum_{i=1}^{n}\frac{a_i}{i+n}=0$$ I tried using $\lim_{n \to \infty}\sum_{i=1}^{n}\frac{a_i}{n}$ as a bound so that I can use comparison test but I can't prove that it converges either. How would I go about proving this statement?
 A: Note that for any $\epsilon > 0$, there exists $N$ such that
$$0 \leqslant \sum_{i=1}^n \frac{a_i}{i+n} = \sum_{i=1}^N \frac{a_i}{i+n}+ \sum_{i=N+1}^n \frac{a_i}{i+n} \leqslant  \sum_{i=1}^N \frac{a_i}{i+n}+ \sum_{i=N+1}^n \frac{a_i}{i} \\ <  \sum_{i=1}^N \frac{a_i}{i+n} + \frac{\epsilon}{2}$$
Try to finish from here.
A: For any natural $K$, you can write
$$0\leq\sum_{i=1}^n\frac{a_i}{i+n}= \sum_{i=1}^K\frac{a_i}{i+n}+\sum_{i=K+1}^n\frac{a_i}{i+n}\leq \sum_{i=1}^K\frac{a_i}{i+n}+\sum_{i=K+1}^n\frac{a_i}{i}.$$
As $\sum_{i=1}^\infty\frac{a_i}{i}=a<\infty$, [any tail must converge to zero][1], which means that $\lim_{n\to\infty}\sum_{i=K+1}^n\frac{a_i}{i}=0$. Thus, from the above expression,
$$0\leq\lim_{n\to\infty}\sum_{i=1}^n\frac{a_i}{i+n}\leq \lim_{n\to \infty}\sum_{i=1}^K\frac{a_i}{i+n}+\lim_{n\to \infty}\sum_{i=K+1}^n\frac{a_i}{i}=\sum_{i=1}^K\lim_{n\to\infty}\frac{a_i}{i+n}=0.$$
[1]: https://proofwiki.org/wiki/Tail_of_Convergent_Series_tends_to_Zero
A: Note that there is a point after which the sum of all terms is below $\epsilon$.
All terms before this point go to zero, obviously, as $\frac{a_i}{i+n}$ converges to zero for any $i$.
Since the $\epsilon$ is chosen arbitrarily, the tail converges to zero.
A other way to look at it, the tail of the series can be as small as desired. Hence as $n$ goes to infinity it should vanish. To prove it rigorously, let $\epsilon \in \mathbb R_+$, now for any $\epsilon$ there is $k_1$
$$\sum_{i=k}^n \frac{a_i}{i+n}<\sum_{i=k}^n \frac{a_i}{i}<\frac{\epsilon}{2}, \space \space k_1<n$$
and
$$\sum_{i=1}^k \frac{a_i}{i}<C$$
So that there is $k_2$
$$\sum_{i=1}^k \frac{a_i}{i+n}<\frac{\epsilon}{2}, \space \space \space k_2<n,$$
then choose $k_3=max(k_1,k_2)$.
