# Prove or disprove if $\sum_{n=1}^{\infty} a_n$ converges, then $\sum_{n=1}^{\infty} \frac{\left | a_n \right |}{n}$ also converges

I'm homelearning calculus and trying to prove or disprove the following problem:

If $$\sum_{n=1}^{\infty} a_n$$ converges, then $$\sum_{n=1}^{\infty} \frac{\left | a_n \right |}{n}$$ also converges.

However, I'm not sure how I would prove or disprove the statement. Could you help me?

Thanks

It doesnt'work, take $$a_n=\frac{(-1)^n}{\log n}$$, then $$\sum a_n$$ converges but $$\sum\frac{|a_n|}{n}=\sum\frac{1}{n\log n}$$ diverges.

• Nice, I have a doubt: the series you've shown isn't defined for $n=1$, but maybe it doesn't matter because if we subtract a finite number of terms from a a series we don't affect its behaviour (we only affect the sum). Is this the reason why your counterexample works even if it starts from $n=2$? Or maybe we can just consider $b_n=\frac{(-1)^n}{\ln (n+1)}$. Thanks. Apr 19 '20 at 15:30
• Yes, the convergence of $\sum_{n\geqslant m}u_n$ is equivalent to the convergence of $\sum_{n\geqslant p}u_n$ for all $m,p$. $b_n$ works aswell if you want to start your sum at $n=1$. Apr 19 '20 at 15:34

If a series is absolutely convergent, then it is convergent. But converse is not true. Because $$\ \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}$$ is convergent but $$\ \sum_{n=1}^{\infty} \dfrac{{1}}{n}$$ is not.

• Your divergent series is $\sum|a_n|$, not $\sum{|a_n|\over n}$. Apr 19 '20 at 21:02