# Does $\sum_{n=1}^\infty \frac {\left(\cos\left(\frac{n\pi}{8}\right)\right)^n}{n}$ converge?

The Problem is: Does the series $$\sum_{n=1}^\infty \frac {\left(\cos\left(\frac{n\pi}{8}\right)\right)^n}{n}$$ converge or diverge or oscillate ?

My approach : Actually, the series is not absolutely convergent because the subsequence $$T_{16k} \gt \sum_{j=1}^k \frac{1}{16j}$$ and hence becomes divergent where $$T_n$$ represents the partial sums of the absolute series .

Now, I have tried to apply the Dirichlet's test on the series but I think the series $$\sum_{n=1}^\infty \left(\cos\left(\frac{n\pi}{8}\right)\right)^n$$ is not bounded because if $$n =8k,8k+1,...,8k+7$$ for all non-negative integers $$k$$ , then except $$n=8k$$, the rest of the series forms into distinct brackets of infinite geometric progressions with common ratio(s) $$r \lt 1$$ but we get $$1+1+..1+...$$ in the series for $$n=8k$$ values and hence the series becomes unbounded .

I can't approach further, a small hint is warmly appreciated .

• What's to the $n$-th power? The cosine or the arc? Commented Jun 12, 2020 at 9:53

Let $$a_n$$ be the $$n$$th term of our series. I'll show $$\sum a_n$$ is the sum of a convergent series and a divergent series. This implies $$\sum a_n$$ diverges.

Define $$b_n=0$$ if $$n$$ is a multiple of $$8,$$ $$b_n=a_n$$ otherwise. Define $$c_n = \frac{1}{n}$$ if $$n$$ is a multiple of $$8,$$ $$c_n=0$$ otherwise. Then $$a_n=b_n+c_n$$ for all $$n.$$

Now $$|b_n| \le (\cos(\pi/8))^n$$ for all $$n,$$ and $$\sum_{n=1}^{\infty}(\cos(\pi/8))^n$$ is a convergent geometric series. Thus $$\sum b_n$$ converges.

As for $$\sum c_n,$$ let $$S_n$$ be its $$n$$th partial sum. Then

$$S_{8m} = \sum_{k=1}^{m} \frac{1}{8k}$$

and thus $$S_{8m}\to \infty$$ as $$m\to \infty.$$ Therefore $$\sum c_n$$ is divergent.

So $$\sum a_n$$ is the sum of a convergent series and a divergent series as advertised, and we're done.