A non cauchy-schwarz approach to the question: Prove that if $\sum{a_k^2}$ converges then $\sum{a_k/k}$ converges.

Question

I wanted to know how to solve this question without the use of the cauchy-schwarz inequality and using more standard methods/tests (such as ratio test, comparison test etc.) for convergence/divergence if possible. Thanks in advance.

Answer 1

Hint: (assuming all $a_k \in \mathbb{R}$)

Just note that according to Cauchy-Schwarz inequality you have $$\left|\sum_{k=1}^N \frac{a_k}{k} \right| \leq \sqrt{\sum_{k=1}^N\frac{1}{k^2}}\cdot \sqrt{\sum_{k=1}^N a_k^2}$$

Edit after OP changed the question to "non-Cauchy-Schwarz":

You may use the inequality GM-QM (geometric/quadratic mean) or you derive directly that

• $|a_k\cdot\frac{1}{k}| \leq \frac{a_k^2 + \frac{1}{k^2}}{2}$

$$\Rightarrow \left|\sum_{k=1}^N \frac{a_k}{k} \right| \leq \sum_{k=1}^N \left|\frac{a_k}{k} \right| \leq \frac{1}{2}\left( \sum_{k=1}^Na_k^2+ \sum_{k=1}^N \frac{1}{k^2} \right)$$

Now the required result follows by comparison.

Answer 2

Using Cauchy–Schwarz inequality, you have

$$\left(\sum_{k=1}^n \frac{a_k}{k} \right)^2 \le \sum_{k=1}^n a_k^2 \sum_{k=1}^n 1/k^2$$

and both series on the right side converge.