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Let $x \in \mathbb{R}$. I have to study the convergence of the following three series:

    1. $$\sum_{n=1}^\infty (\sin(\sin n))^n$$
    1. $$\sum_{n=1}^\infty \frac{n \sin (x^n)}{n + x^{2n}}$$
    1. $$\sum_{n=1}^\infty \frac{ \sin (x^n)}{(1+x)^n} $$

I'm completely stumped by the first one. For the second and the third I've tried to study the absolute convergence by using root and ratio tests, but with no good results. I've also try to compare the series number 3 with $\sum_{n=1}^\infty \frac{1}{(|1+x|)^n}$, but again no results.

Could you help me?

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  • $\begingroup$ @Farnight Yes: $x \in \mathbb{R}$ $\endgroup$ – mathlearner May 20 '15 at 10:50
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Hints

For (1), we have $\sin n \in [-1, 1]$ and since $\sin$ is increasing on $[-1, 1]$ we have $\sin \sin n \in [-\sin 1, \sin 1]$ (note too that $0 \leq \sin 1 \leq 1$).

For (2), applying the Comparison and Ratio Tests gives that the series converges for $|x| > 1$ (noting that for particular negative values of $x$ certain terms in the series do not exist for division-by-zero reasons).

For (3), as in (2), again applying the Comparison and Ratio Tests gives converges for a certain set of $x$ values.

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  • $\begingroup$ Then, is it sure that the root test is satisfied (that is, $\sin \sin n$ is strictly less than $1$)? If yes (as it seems) why? $\endgroup$ – mathlearner May 20 '15 at 10:48
  • $\begingroup$ Could you please add more details? Unfortunately, as the answer is, it gives me no help since I've actually already tried these tests with no results. $\endgroup$ – mathlearner May 20 '15 at 10:52
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    $\begingroup$ It's not enough that $\sin \sin n$ is strictly less than $n$ for each $n$ (it's possible for a sequence to have ratios all less than $1$ but converging to $1$); it is enough that $|\sin \sin n| \leq C$ for some constant $C < 1$ for all $n$. $\endgroup$ – Travis Willse May 20 '15 at 10:54
  • $\begingroup$ For (1), $|\sin \sin n| \leq \sin 1$, so the series is bounded by $\sum_{n = 1}^{\infty} (\sin 1)^n$, and this latter series converges by the Ratio Test (the ratio is $\sin 1 < 1$). $\endgroup$ – Travis Willse May 20 '15 at 10:56
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    $\begingroup$ @mathlearner That's because $0<1<\pi/2$ $\endgroup$ – Kitegi May 20 '15 at 11:06
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The second one is simple, if $|x|<1$, use the fact that: $\sin x^n\approx x^n$. And if $|x|>1$, then $|\sin x^n|<1$ and you only need to look at the denominator


For the third one, we have $$\sum \frac{\sin(x^n)}{(1+x)^n}\leq\sum \frac{1}{|1+x|^n}<\infty$$ And $$\sum \frac{\sin(x^n)}{(1+x)^n}\leq\sum \left|\frac{x}{1+x}\right|^n<\infty$$

So if $|1+x|>1$ or $\left|\frac{1+x}{x}\right|>1$, then the sum converges. This gives us that the sum converges for $x>-1/2$ and for $x<-2$.
For $-1<x<-1/2$: $$\frac{\sin(x^n)}{(1+x)^n}\approx\left(\frac{x}{1+x}\right)^n \tag{$\sin(x)\approx_{x\to 0} x$}$$

which doesn't tend to $0$, so the series diverges ($\left|\frac{x}{1+x}\right|>1$).

The only remaining case is $-2\leq x<-1$. In this case $|1+x|\leq 1$. So we only need to show that $\sin x^n$ doesn't tend to $0$, when $1<x<2$. But this doesn't seem like an easy task. I'll post another answer if I come up with anything.

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