does $\sum_{n=1}^\infty \frac {1}{n^{1.7 +\sin n}}$ converge? $$\sum_{n=1}^\infty \frac {1}{n^{1.7 + \sin(n)}}$$
i was trying resolve  it , by any method convergence , but I could not show if the series converge or diverge
 A: Let's try to look at the set $\{ n \in \mathbb{N} : sin(n) < -0.8 \}$. It is easy to see that in general $sin(x)<-0.8$ if 
$$- 0.643+ 3\pi/2 + 2k\pi<x< 0.643 + 3 \pi /2+2k\pi;$$
in particular the intervals where this holds is always longer than $2$ and so for every $k$ there is a natural number $n_k$ that sarisfies this inequality; in particular we have $2k\pi<n_k<2(k+1)\pi$ and $sin(n_k) < -0.8$. Coming back to our sum we can consider only these integers:
$$\sum_{n=1}^{\infty} \frac 1{ n^{1.7 + sin(n)}} \geq \sum_{k=1}^{\infty} \frac 1{n_k^{1.7 + sin(n_k)}} \geq \sum_{k=1}^{\infty} \frac 1{{(2(k+1)\pi)}^{0.9}} = \frac 1{(2\pi)^{0.9}} \sum_{k=1}^{\infty} \frac 1{(k+1)^{0.9}} $$
and the last series diverges by comparison with the harmonic series.
Notice that this proof works also for $2-\epsilon$ in place of $1.7$ as long as the interval where $sin(x)<-(1-\epsilon)$ is longer than $1$. But still the sum diverges for every $\epsilon>0$, looking to the density of the set $\{ sin (n) \}$ in $[-1,1]$.
