# Study the convergence of the series: $\sum_{n=2}^{\infty} \frac{\cos(nx)\sin(\frac{x}{n})}{\ln n}$

I am studying the convergence of the following series:

$$\sum_{n=2}^{\infty}\frac{\cos(nx)\sin\frac{x}{n}}{\ln n}$$

I thought about using the Dirichlet's Test, according to which:

If we have a series of the form $$\sum_{n=0}^{\infty}a_nb_n$$ with

$$\,\,(i)\,\, a_n$$ is decreasing and tends to $$0$$, and

$$(ii)\,\, t_n = b_0 + b_1 +\cdots+ b_n$$ is bounded,

then $$\sum_{n=0}^{\infty}a_nb_n$$ is convergent.

Now, my only problem is that I do not really know how to choose $$a_n$$ and $$b_n$$.

I though about choose $$a_n = ln\space n$$ and $$b_n=\cos(nx)\space \sin\frac{x}{n}$$ but I do not really know if I can calculate the sum for $$b_n$$.

Can you help me find out how to solve this?

## 1 Answer

A. If $$x\ne 2k\pi$$, then, $$c_n=\sin\Big(\frac{x}{n}\Big)$$ eventually maintains sign, say $$\sigma$$ is its eventual sign, and its absolute value is strictly decreasing and tends to zero. In particular $$b_n=\frac{\sigma\sin\big(\frac{x}{n}\big)}{\ln n},$$ is eventually strictly decreasing and tends to zero.

Meanwhile, $$A_n=\cos x+\cos 2x+\cdots+\cos nx=\mathrm{Re}\big(\mathrm{e}^{xi}+ \mathrm{e}^{2xi}+\cdots+\mathrm{e}^{nxi}\big) =\mathrm{Re}\left(\frac{\mathrm{e}^{(n+1)xi}-\mathrm{e}^{xi}}{\mathrm{e}^{xi}-1}\right)$$ and thus $$|A_n|\le \frac{2}{|\mathrm{e}^{xi}-1|}=\frac{2}{|\sin(x/2)|},$$ and hence $$A_n$$ is bounded, whenever $$x\ne2k\pi$$. Then convergence is a consequence of Dirichlet's Test.

B. If $$x=2k\pi$$ and $$x\ne 0$$, then our series looks like $$\sum_{n=2}\frac{\sin (\frac{x}{n})}{\ln n}$$ in which case we obtain divergence via Limit Comparison Test, since $$\frac{\displaystyle\frac{\sin (\frac{x}{n})}{\ln n}}{\displaystyle\frac{1}{n\ln n}}=n\sin\big(\frac{x}{n}\big)\to x.$$

C. If $$x=0$$, then the sum is equal to zero.

• Isn't the series divergent if $x=2k\pi\ne0$? – Learner Nov 4 '18 at 10:07
• @Learner Correct! – Yiorgos S. Smyrlis Nov 4 '18 at 10:08
• Ah perfect, nice answer – Learner Nov 4 '18 at 10:11
• Boundedness of $A_n$ is not quite trivial… – Bernard Nov 4 '18 at 12:19
• How do you know that $c_n$ is decreasing? – Ghost Nov 6 '18 at 8:33