# Conditional convergence of series $\sum_{n=1}^{\infty}a_n(\frac{k}{n}-k^{n+1}\frac{\zeta(n+1)}{n+1}),0<k\le 1$

Does the series $$\displaystyle \sum_{n=1}^{\infty} a_n\left(\frac{k}{n}-k^{n+1}\frac{\zeta(n+1)}{n+1}\right),0 converges conditionally given that $$\sum_{n=1}^{\infty}|a_n|<\infty$$.

What I know is that this series converges . Setting $$a_n=\frac{1}{2^n}$$ and $$k=0.5$$, Wolfram alpha shows that the series converges conditionally. So I guess the series might be conditionally convergent for every such $$k$$ and $$a_n$$ but I have no idea how to prove it. Any help/hint is appreciated.

EDIT: I tried to use some convergence tests from here. Mathematica was unable to compute the limits to apply Root and Ratio test. Also applying the first Summand test, the limit turns out to be $$0$$ and hence inconclusive.

• Replacing $\zeta(n+1)$ by $1$ won't change anything – reuns Mar 16 at 18:41

It seems that the series converges absolutely:

$$\left|\frac{k}{n}-k^{n+1}\frac{\zeta(n+1)}{n+1}\right| \le \frac{1}{n} +\frac{1}{n+1}\zeta(2) \le 1+\zeta (2).$$

Thus

$$\sum_{n=1}^{\infty} |a_n|\left|\frac{k}{n}-k^{n+1}\frac{\zeta(n+1)}{n+1}\right| \le \sum_{n=1}^{\infty}|a_n|(1+\zeta(2))<\infty.$$

• @ zhw.This is what I have tried initially. Your approach is essentially using the triangle inequality to prove absolute convergence whereas absolute convergence is necessary to apply triangle inequality itself. The discussion is here. – ersh Mar 16 at 19:53
• I'm not sure why you are citing that discussion. I've shown your series converges absolutely. Therefore it converges. – zhw. Mar 16 at 20:07
• I cited it to mention that we cannot apply triangle inequality to infinite series without the condition that the series should be absolutely convergent. Speaking precisely, if $\sum a_n$ is given series then $\mid\sum a_n\mid\le\sum\mid a_n\mid$ is not true in general unless we already know that series $\sum\mid a_n\mid$ converges. If you see your argument, you are using triangle inequality to bound the terms of the series, which we can only use if we already know that the series $\sum\mid a_n\mid$ is convergent. – ersh Mar 16 at 20:21
• Also in my original post, Wolfram alpha shows that the series diverges for specific values of $a_n$ and $k$, I don't doubt wolfram alpha :D – ersh Mar 16 at 20:23
• Actually, zhw. is correct!. I got overwhelmed by wolfram alpha and the application of triangle inequality in infinite case. zhw.'s answer shows that the series converges absolutely and uniformly by Weirstrass test. My bad..this is so basic. Anyway, I didn't know Wolfram Alpha could be wrong! – ersh Mar 16 at 22:27