Context

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.

Doubt

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.

  • Yes, your alternate technique is perfectly valid: the term-by-term sum of three convergent series is itself convergent. – Greg Martin Sep 17 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 $$\begin{eqnarray*} \sum_{n\geq 1}\frac{H_n}{n}\sin(2n) &=& \sum_{m\geq 1}\frac{\sin(2m)}{m^2}+\frac{1}{2}\text{Im}\left[\log(2\sin 1)-i\left(\frac{\pi}{2}-1\right)\right]^2\\&=&-\left(\frac{\pi}{2}-1\right)\log(2\sin 1)+\sum_{m\geq 1}\frac{\sin(2m)}{m^2}\approx 0.43.\end{eqnarray*} $$

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.