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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?

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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.

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

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