# Prove that $\sum_{n=0}^{+\infty}\frac{1}{n^{1+\cos(n)}}$ diverges

Prove that $$\sum_{n=0}^{+\infty}\frac{1}{n^{1+\cos(n)}}$$ diverges.

At first I consider the set $$P:=\{\lfloor(2n+1)\pi \rfloor: n\in\mathbb{N}\}$$.

So the series $$\sum_{n=0}^{+\infty}\frac{1}{n^{1+\cos(n)}}\ge\sum_{n\in P}\frac{1}{n^{1+\cos(n)}}=\sum_{n=0}^{+\infty}\frac{1}{n^{1+\cos(\lfloor(2n+1)\pi \rfloor)}}$$

$$\cos(\lfloor(2n+1)\pi \rfloor)=\cos((2n+1)\pi-\{(2n+1)\pi\})$$, where $$\{(2n+1)\pi\}$$ is the fractional part. So using a trigonometric formula I have that $$\cos(\lfloor(2n+1)\pi \rfloor)=-\cos(\{(2n+1)\pi\})$$.

I suppose that the set $$\{\{(2n+1)\pi\}:n\in\mathbb{N}\}$$ is dense in $$[0,1]$$, because i know that for every $$x$$ irrational the set $$\{\{nx\}:n\in\mathbb{N}^*\}$$ is dense in $$[0,1]$$.

So exists a succession $$\alpha_n$$ in $$\{\{(2n+1)\pi\}:n\in\mathbb{N}\}$$ such that $$\alpha_n\rightarrow0$$ when $$n$$ goes to $$\infty$$.

For definition of limit I have that $$\forall\epsilon>0,\exists N>0$$ such that $$\forall n>N$$ I have that $$|\alpha_n|<\epsilon$$.

So I estimate the series from below whit

$$\sum_{n=0}^{+\infty}\frac{1}{n^{1-\cos(\alpha_n)}}\ge\sum_{n>N}\frac{1}{n^{1-\cos(\epsilon)}}$$

For arbitrariness of $$\epsilon$$, when $$\epsilon\rightarrow0$$ I have that $$1-\cos(\epsilon)\sim\frac{\epsilon^2}{2}$$ and $$n^{1-\cos(\epsilon)}\sim n^{\frac{\epsilon^2}{2}}\sim 1$$ So the series $$\sum_{n>N}\frac{1}{n^{1-\cos(\epsilon)}}$$ diverges.

I do not know my reasoning is right, because I have the following hint in the text:

$$P:=\{\lfloor(2n+1)\pi \rfloor: n\in\mathbb{N}\}$$ has natural density strictly positive. Where natural density is $$d(P)=\lim_n\frac{|P\cap[1,n]|}{n}$$

• Yes, $\alpha_n$ is in the set of fractional part $\{(2n+1)π\}_n$. – Simmetrico Mar 6 at 16:21

$$\cos(n) = \cos(n - 2 k \pi)$$ where $$k = \lfloor n/(2\pi) \rfloor$$, and $$n - 2k\pi = 2\pi \{n/(2\pi)\}$$. Whenever $$\{n/(2\pi)\} \in (1/3, 2/3)$$, we have $$\cos(n) < -1/2$$. And since the set of such $$n$$ has positive density...
• If more than fraction $r$ of the integers from $N$ to $2N-1$ are in the set, then $$\sum_{n=N}^{2N-1} \frac{1}{n^{1+\cos(n)}} \ge \frac{r N}{(2N)^{1/2}}$$ – Robert Israel Mar 7 at 2:08
Let $$(n_r)_{r \ge 1}$$ be a sequence in $$\mathbb{N}$$ such that $$n_r \in [ \pi/3 + 2 \pi r , 2 \pi/3 + 2 \pi r ]$$ for all $$r \in \mathbb{N}$$. Such an integer $$n_r$$ always exists since $$\pi/3 > 1$$. Then
$$\sum_{n=1}^{\infty} \frac{1}{n^{1+cos(n)}} \ge \sum_{r=1}^{\infty} n_r^{-1/2} \ge \sum_{r=1}^{\infty} (2 \pi r )^{-1/2}$$