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$$\sum_{n=1}^\infty \frac {1}{n^{1.7 + \sin(n)}}$$

i was trying resolve it , by any method convergence , but I could not show if the series converge or diverge

Antonio Vargas
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  • Do add a dollar sign "$" before and after a mathematical expression – Timbuc Sep 04 '14 at 16:17
  • http://math.stackexchange.com/questions/6818/testing-the-series-sum-limits-n-1-infty-frac1nk-cosn/6824#6824 –  Sep 04 '14 at 22:25

1 Answers1

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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 $[-1,1]$.

StheW
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  • Shouldn't we be concerned with $\sin(n) \leq -0.7$? – graydad Sep 04 '14 at 16:44
  • @graydad: exactly. Here it is shown that there are "many" $n$ for which $\sin(n)\leq -0.8$ (even worse that $\sin(n)\leq -0.7$). So many that their contribute makes the sum diverge. – Jack D'Aurizio Sep 04 '14 at 17:03
  • Technically, the density of ${\sin n}$ isn't enough to show that the sum diverges for all $\epsilon\gt 0$ - you need something along the lines of equidistribution for this. (There are dense sequences which aren't equidistributed) – Steven Stadnicki Sep 04 '14 at 17:21
  • Ok, to be clearer, taking in account the continuity of ${\rm sin}(x)$, it is sufficient is that for every $\epsilon >0$ we have the existence of $N_{\epsilon}$ such that in every set of the type ${ j, j+1, j+2, \ldots, j+N_{\epsilon}}$ there is an integer $n$ such that $|n-(2k+3/2)\pi| \leq \epsilon$ for some integer $k$. But this is true since $\pi$ is irrational. ciao jack, mi hai contagiato!! – StheW Sep 05 '14 at 08:36