Does such a convergent series exist?

Question

This is the question that arose in my mind while contemplating the convergence of this series: $$\sum_{n\geq 1} \frac{\sin(\sqrt{n})}{n}$$

In this problem, if we assume $A_N = \sum_{n=1}^{N}\sin(\sqrt{n})$, and consider the method of partial sums:

$$\sum_{n=1}^{N} \frac{\sin(\sqrt{n})}{n} = \frac{A_N}{N}+\sum_{n=1}^{N-1}\frac{A_n}{n(n+1)}$$

Then, due to the estimation of $A_N = O(\sqrt{N})$, we know that the series converges. At this point, I have a question regarding the summation of general form like $\sum_{n\geq 1} \frac{a_n}{n}$, we can make the same assumption that $A_N = \sum_{n=1}^{N}a_n$, and apply the method of partial summation $$\sum_{n=1}^{N} \frac{a_n}{n} = \frac{A_N}{N}+\sum_{n=1}^{N-1}\frac{A_n}{n(n+1)}$$

In this case, if $|A_N| \gg N$, then this method cannot determine the convergence of the series; it can only conclude that the series is not absolutely convergent. For example, let $A_N = (-1)^N N$, in this case, the series does not converges. However, does these exist $|A_N| \gg N$ such that the series $\sum_{n=1}^{\infty} \frac{a_n}{n}$ conditionally converges?

Answer 1

I don't believe so. Assume that $|A_N| \gg N$.

• If $\{A_N\}$ changes sign infinitely often, then $|a_n| = |A_n - A_{n-1}| = |A_n| + |A_{n-1}| \gg n$ infinitely often, so that $\frac{a_n}n$ does not tend to $0$.
• Otherwise $\{A_N\}$ is eventually of constant sign (without loss of generality, always positive), in which case $\sum_{n=1}^{N-1} \frac{A_n}{n(n+1)} \gg \sum_{n=1}^{N-1} \frac1{n+1} \gg \log N$ diverges.