I am trying to show that the series(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\sum_{n=1}^{\infty}\frac{(\frac{2}{3}+\frac{1}{3}\sin{n})^n}{n}[/tex] converges (it's a variant of the harmonic series).

So far I got

1) The sequence of partial sums is monotone increasing

2)

[tex]\frac{1}{3}\leq\frac{2}{3}+\frac{1}{3}\sin{n}\leq1[/tex]

[tex]\frac{1}{3^n}\leq(\frac{2}{3}+\frac{1}{3}\sin{n})^n\leq1^n[/tex]

[tex]0<(\frac{2}{3}+\frac{1}{3}\sin{n})^n\leq1[/tex]

Since

[tex]\sin{x}=1\Rightarrow{x}=\frac{\pi}{2}+2m\pi[/tex]

and [tex]\pi[/tex] is irrational

[tex]\sin{n}<1[/tex]

for all rational or integraln; thus

[tex]0<(\frac{2}{3}+\frac{1}{3}\sin{n})^n<1\ \forall{n}\in\mathbb{N}[/tex].

What I've been trying to do here is prove that the sequence of partial sums is Cauchy (edit: which is sufficient to show that it is bounded and the series converges):

[tex]\sum_{n=j}^{k}\frac{(\frac{2}{3}+\frac{1}{3}\sin{n})^n}{n}<\epsilon[/tex]

But I can't figure out how to show that [tex]\sin{n}[/tex] is not arbitrarily close to 1 at some point, and that

[tex](\frac{2}{3}+\frac{1}{3}\sin{n})^n[/tex]

is itself not arbitrarily close to 1; thus hindering attempts to make the sum vanish.

I also tried using the power series expansion of [tex]\sin{n}[/tex] (which converges everywhere):

[tex](\frac{2}{3}+\frac{1}{3}\sin{n})^n=(\frac{2}{3}+\frac{1}{3}(n-\frac{n^3}{3!}+\frac{n^5}{5!}-...))^n=(\frac{2}{3})^n+n(\frac{2}{3})^{n-1}(\frac{n}{3}-\frac{n^3}{3\cdot3!}+...)+...[/tex]

(assuming this is valid, I don't know where it leads)

So what's the trick?

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# Difficult question - convergence of a series

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