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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 $$d(P)=\lim_n\frac{|P\cap[1,n]|}{n}$$

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

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  • $\begingroup$ How I can use the positive density? $\endgroup$ – Simmetrico Mar 6 at 20:56
  • $\begingroup$ If more than fraction $r$ of the integers from $N$ to $2N-1$ are in the set, then $$\sum_{n=N}^{2N-1} \frac{1}{n^{1+\cos(n)}} \ge \frac{r N}{(2N)^{1/2}}$$ $\endgroup$ – Robert Israel Mar 7 at 2:08
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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

$$ \sum_{n=1}^{\infty} \frac{1}{n^{1+cos(n)}} \ge \sum_{r=1}^{\infty} n_r^{-1/2} \ge \sum_{r=1}^{\infty} (2 \pi r )^{-1/2}$$

which indeed diverges.

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