Convergence of $\lim_{n \to \infty}\sum_{i=1}^{n}\frac{a_i}{i+n}$

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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 $k_3=max(k_1,k_2)$.