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$$\sum_{n=1}^\infty (\sin{\frac{1}{n}})(\sinh{\frac{1}{n}})(\cos n)$$

I tried this

$(\sin{\frac{1}{n}})(\sinh{\frac{1}{n}})(\cos n)\le \sinh{\frac{1}{n}}$ and then i want to prove that $\sinh{\frac{1}{n}}$ converges. I am kind of stuck at this point.

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  • $\begingroup$ $$\sum_{n=1}^\infty (\sin{\frac{1}{n}})(\sinh{\frac{1}{n}})(\cos n) \approx 0.318223$$ - it converges $\endgroup$ – Yuriy S Jun 11 '16 at 11:18
  • $\begingroup$ @YuriyS This does not address the question, does it? $\endgroup$ – Did Jun 11 '16 at 12:29
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Because

$$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=\lim_{x\rightarrow 0} \frac{\sinh(x)}{x}=1$$

thus

$$\lvert \sin \left( \frac{1}{n} \right)\sinh \left( \frac{1}{n} \right)\cos(n)\rvert \leq \lvert \sin \left( \frac{1}{n} \right) \sinh \left( \frac{1}{n} \right) \rvert \simeq \frac{1}{n^2}$$

Prove that $\lim\limits_{x \to 0} \sinh(x)/x =1$.

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    $\begingroup$ Shouldn't we prove that $\sin{\frac{1}{n}}\sinh{\frac{1}{n}} \le \frac{1}{n^2}$ $\endgroup$ – Rush ThaMan Jun 11 '16 at 11:38
  • $\begingroup$ @RushThaMan, not necessarily. $$\sum_{n \geq 1} \frac{1}{n^s}$$ converges for any $s >1$ $\endgroup$ – Yuriy S Jun 11 '16 at 11:55
  • $\begingroup$ @YuriyS Irrelevant, only the series $\sum\frac1{n^2}$ is needed here. $\endgroup$ – Did Jun 11 '16 at 12:33
  • $\begingroup$ @RushThaMan No we do not have to, actually, $0\leqslant\sin(1/n)\sinh(1/n)\leqslant42/n^2$ suffices. $\endgroup$ – Did Jun 11 '16 at 12:33
  • $\begingroup$ @Did why would $\sin{1/n} \sinh{1/n}$ be less than $42/n^2$ $\endgroup$ – Rush ThaMan Jun 11 '16 at 12:47
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You can partition the set $\mathbb{N}\bigcap[1,\infty)$ into a countable set of sets $\{A_1, A_2, ... \}$, where the cardinality of $A_i$ is less than 5, and $A_i$ contains consecutive integers only, and such that $n_1,n_2\in A_i \Rightarrow\text{sign}(\cos(n_1)) =\text{sign}(\cos(n_2))$.

Then because of the finite cardinality and convergence towards zero of the terms in your sum

$$\lim_{i\to\infty}\sum_{n\in A_i}(\sin{\frac{1}{n}})(\sinh{\frac{1}{n}})\cos(n)=0$$

The following two sums are the same because of the consecutiveness in the $A_i$ sets, and the first is alternating and has terms converging to zero which makes it convergent:

$$\infty>\sum_{i=1}^{\infty}\sum_{n\in A_i}(\sin{\frac{1}{n}})(\sinh{\frac{1}{n}})\cos(n)=\sum_{n=1}^{\infty}(\sin{\frac{1}{n}})(\sinh{\frac{1}{n}})\cos(n)$$

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  • $\begingroup$ This would address the convergence of $\sum\sin(1/n)\cos (n)$ and $\sum\sinh(1/n)\cos (n)$ but for $\sum\sin(1/n)\sinh(1/n)\cos (n)$, such delicacy is not needed. $\endgroup$ – Did Jun 11 '16 at 12:30
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It converges absolutely because $$\biggl\lvert\sin\frac1n\cdot\sinh\frac1n\cdot\cos n\biggr\rvert=_\infty O\biggl(\frac 1{n^2}\biggr),$$ and the series $\displaystyle\sum_{n\ge 1}\frac1{n^2}\;$ converges.

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