# If $\sum |a_k|$ converges, then $\sum |a_k|/{k^p}$ converges for all $p\ge 0$

Problem: If $\sum_{k=1}^{\infty} |a_k|$ converges, prove that $\sum_{k=1}^{\infty} \frac{|a_k|}{k^p}$ converges for all $p\geq0$. What happens if $p<0?$

Attempt: Suppose $\lim_{k→\infty}\frac{|a_k|}{k^p} = 0$ exists for all p greater than zero. Also given that $\sum_{k=1}^{\infty} \frac{|a_k|}{k^p}$ converges for all $p\geq0$, we have by the limit comparison test $\sum_{k=1}^{\infty} \frac{|a_k|}{k^p}$ is convergent when $p\geq 0$. If $p<0$ then $\sum_{k=1}^{\infty} \frac{|a_k|}{k^p} = \infty$. Hence $\sum_{k=1}^{\infty} \frac{|a_k|}{k^p}$ is divergent when p < 0.

• Checking a few simple, concrete, cases might help you to eliminate the desire to prove some dubious assertions... – Did Feb 12 '15 at 9:31
• what type of concrete cases? – user848242 Feb 12 '15 at 9:33
• I can't quite understand your argument for $p\geq 0$ case. Can you elaborate? – user99914 Feb 12 '15 at 9:35
• So it is convergent still convergent when p < 1. Is the first part ok? – user848242 Feb 12 '15 at 9:35
• "what type of concrete cases?" Tell us! On which simple examples of series do you test assertions about the convergence of series? – Did Feb 12 '15 at 9:38

If $p > 0$ then $k \ge 1$ implies $k^p \ge 1$. Thus $\dfrac{|a_k|}{k^p} \le |a_k|$ for all $k$ and the series $\displaystyle \sum_{k=1}^\infty |a_k|$ converges by the comparison test.
Your statement if $p < 0$ is false. $\displaystyle \sum_{k = 1}^\infty k^{-3}$ converges, but so does $\displaystyle \sum_{k=1}^\infty \frac{k^{-3}}{k^{-1}}=\sum_{k = 1}^\infty k^{-2}$
Case $p>0: \ \exists n_0 >n \ \forall n$ s.t. $|a_n| = b_{n \geq n_0} < 1$ - this is the criteria for convergence. Clearly if $p>0 \ b_n >\frac{b_n}{n^p} = c_n$, therefore $\sum_n c_n$ converges by comparison test.
Case $p<0$: take e.g. $b_n = \frac{1}{n^{1+\frac{1}{2}}}$ and $p=-\frac{1}{2}$. What do you get?