# Convergence of a series

[SOLVED] convergence of a series

## Homework Statement

Prove that the convergence of $\sum a_n$ implies the convergence of $\sum \frac{a_n}{n}$ if $a_n \geq 0$.

## The Attempt at a Solution

I want to use the comparison test. So, I want to find $N_0$ so that $n \geq N_0$ implies $$\frac{\sqrt{a_n}}{n} \leq a_n$$ which is clearly not in general possible. So I am stuck. Maybe I need to replace a_n by a subsequence of a_n or something?

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consider any partial sum A_m = Sum(a_k/k, k = 1,..., m).

|A_m| = |a_1/1 + ... + a_k/k| = |(1/1) * a_1 + ... + (1/k)*a_k| <= ...

Sorry. I am trying to prove
$$\sum \frac{\sqrt{a_n}}{n}$$

converges not

$$\sum \frac{a_n}{n}$$

For some reason, I cannot edit the opening post.

ok even easier, just means you won't have to show something else I had in mine,

Let A_m = sum(sqrt(a_k)/k, k = 1,..., m) be any partial sum, then

|A_m| = |sqrt(a_1)/1 + ... + sqrt(a_k)/k| = |(1/1)*sqrt(a_1) + ... + (1/k)sqrt(a_k)| <= ...

use some famous inequality for the next step(not the triangle inequality), what's the other one I'm sure you know!

Cauchy-Schwarz. Yay!

BTW you're k's and m's are mixed up.

Cauchy-Schwarz. Yay!

BTW you're k's and m's are mixed up.
woops yea, good to see you got it! I'll look at your other question now