Does the series ∞∑n=11n1+|sin(n)| \sum\limits_{n=1}^{\infty} \frac{1}{n^{1 + |\sin(n)|}} converge or diverge?

Does the following series converge or diverge? I would like to see a demonstration.

n=11n1+|sin(n)|.

I can see that:
n=11n1+|sin(n)|

Answer

This is problem 11162, posed by Paolo Perfetti, in the June-July 2005 issue
of the American Mathematical Monthly. The solution below, due to the Microsoft Research
Problems Group, is found in the February 2007 issue of the same magazine.

For positive integer n, define
A_n=[0,2^n)\cap\{k\in \mathbb{N}:|\!\sin k|<\textstyle{1\over n}\},\quad B_n=[2^{n-1},2^n)\cap A_n.
If k\in B_n, then k^{-1-|\sin k|}>(2^n)^{-1-1/n }=2^{-n-1}. If n>1, then A_n is
contained in the disjoint union of A_{n-1} and B_n, so |B_n|\geq |A_n|-|A_{n-1}|. To estimate
|A_n|, partition the unit circle into 7n arcs, each with angle 2\pi/(7n).
Of the values e^{ik} for 0\leq k<2^n, at least 2^n/(7n) lie in the same arc by the
Pigeonhole Principle. If e^{ik_1} and e^{ik_2} lie in the same arc, then
|\sin(k_1-k_2)|\leq |e^{i(k_1-k_2)}-1|=|e^{ik_1}-e^{ik_2}|<{2\pi\over7n}<{1\over n}
and |k_1-k_2|\in A_n. Subtracting the smallest k from the others (and itself), we find
that |A_n|\geq2^n/(7n). Now if N\geq2, then
\begin{eqnarray*}
\sum_{k=2}^{2^N-1}k^{-1-|\sin k|}&=&\sum_{n=2}^N\sum_{k=2^{n-1}}^{2^n-1}k^{-1-|\sin k|}
\geq \sum_{n=2}^N\sum_{k\in B_n}k^{-1-|\sin k|}\geq\sum_{n=2}^N {|B_n|\over2^{n+1}}\\[9pt]
&\geq& \sum_{n=2}^N{|A_n|-|A_{n-1}|\over 2^{n+1}}=\sum_{n=2}^N
\left(\left( {|A_n|\over 2^{n+2}}-{|A_{n-1}|\over 2^{n+1}}\right)+{|A_n|\over 2^{n+2}}\right) \\[9pt]
&=&{A_N\over 2^{N+2}}-{|A_1|\over8}+\sum_{n=2}^N{|A_n|\over 2^{n+2}}\\[9pt]
&\geq& -{|A_1|\over8}+\sum_{n=2}^N {2^n/(7n)\over2^{n+2}}= -{|A_1|\over8}+\sum_{n=2}^N {1\over28n}
\end{eqnarray*}

This grows without bound as N\to\infty.

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