I am doing an exercise:

Test convergence of $$ \sum_1^\infty \frac {H_n}n \sin(2n), $$ where $$ H_n = 1 + \frac 12 + \cdots + \frac 1n $$

Obviously we can use the Dirichlet test, since $H_n/n \searrow 0$ and $\sum_1^N \sin(2n)$ is bounded by 2. However, I am inclined to use the following technique:

Use the well-known asymptotic formula $$ H_n = \log(n) + \gamma + O\left( \frac 1n\right) \quad [n\to \infty], $$ then the original series could formally be broken into $$ \sum \frac {\log(n) \sin(2n)}n + \sum \frac {\gamma \sin(2n)}n + \sum \sin(2n) \cdot O\left(\frac 1{n^2}\right), $$ and each of them is convergent [the 1st and the 2nd follows the Dirichlet test, the 3rd is absolutely convergent], then so is the original series.


We know that

when $\sum a_n, \sum b_n$ are series of nonnegative terms and $\sum a_n$ converges, if $a_n \sim b_n$, then $\sum b_n$ is also convergent.

Clearly this generally fails for series with terms of nonconstant signs. But can we do something as I showed above and determine the convergence like that? I think we could do this, but not 100% sure.

  • $\begingroup$ Yes, your alternate technique is perfectly valid: the term-by-term sum of three convergent series is itself convergent. $\endgroup$ – Greg Martin Sep 17 '18 at 6:24

Let us assume $|x|<1$. Since $-\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n}$ we have $$\frac{-\log(1-x)}{1-x} = \sum_{n\geq 1} H_n x^n $$ and by applying $\int(\ldots)\frac{dx}{x}$ to both sides we get: $$ \text{Li}_2(x)+\frac{1}{2}\log^2(1-x) = \sum_{n\geq 1}\frac{H_n}{n}x^n. $$ The wanted series is the imaginary part of the RHS evaluated at $x=e^{2i}$ (summation by parts ensures the convergence at such a point, belonging to the boundary of $|x|<1$), hence $$\begin{eqnarray*} \sum_{n\geq 1}\frac{H_n}{n}\sin(2n) &=& \sum_{m\geq 1}\frac{\sin(2m)}{m^2}+\frac{1}{2}\text{Im}\left[\log(2\sin 1)-i\left(\frac{\pi}{2}-1\right)\right]^2\\&=&-\left(\frac{\pi}{2}-1\right)\log(2\sin 1)+\sum_{m\geq 1}\frac{\sin(2m)}{m^2}\approx 0.43.\end{eqnarray*} $$


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