# convergence of $\sum_{n=1}^{\infty}(n^2+1)\log^\alpha \left ( \frac{sh\frac{1}{n}}{\sin\frac{1}{n}} \right )$

$\sum_{n=1}^{\infty}(n^2+1)\log^\alpha \left ( \frac{sh\frac{1}{n}}{\sin\frac{1}{n}} \right )$

Find the value of $\alpha$ for which the series converges

please, I have no idea to approach the solution thanks.

• "sh" is hyperbolic sine? You should probably take Taylor series and see what the leading term is, and then choose $\alpha$ so that it's $<-1$. – Eric Auld Nov 13 '16 at 20:00
• @EricAuld: Yes, this is standard notation in some parts of the world. – Alex M. Nov 13 '16 at 20:03
• yes, sh(x)=(e^x-e^-x)/2 – Student Nov 13 '16 at 20:03

Hint: use Taylor series, again, again, and again.

In detail:

From a Taylor series around $0$, since $\frac{1}{n}\xrightarrow[n\to\infty]{}0$:

$$\sinh \frac{1}{n} = \frac{1}{n} + \frac{1}{6n^3} + o\left(\frac{1}{n^3}\right) \tag{1}$$ and $$\sin \frac{1}{n} = \frac{1}{n} - \frac{1}{6n^3} + o\left(\frac{1}{n^3}\right)\tag{2}$$ from which $$\frac{\sinh \frac{1}{n}}{\sin \frac{1}{n}} = \frac{1+\frac{1}{6n^2}+ o\left(\frac{1}{n^2}\right)}{1-\frac{1}{6n^2}+ o\left(\frac{1}{n^2}\right)} = 1+\frac{1}{3n^2}+ o\left(\frac{1}{n^2}\right).$$ (Using $\frac{1}{1+x}= 1-x+o(x)$ around $0$). From the Taylor series $\log(1+x)=x+o(x)$ at $0$, we get $$\log^\alpha \frac{\sinh \frac{1}{n}}{\sin \frac{1}{n}} = \log^\alpha\left( 1+\frac{1}{3n^2}+ o\left(\frac{1}{n^2}\right)\right) = \left( \frac{1}{3n^2}+ o\left(\frac{1}{n^2}\right)\right)^\alpha = \frac{1}{3^\alpha n^{2\alpha}}+ o\left(\frac{1}{n^2\alpha}\right)$$ and finally $$(n^2+1)\log^\alpha \frac{\sinh \frac{1}{n}}{\sin \frac{1}{n}} = \frac{1}{3^\alpha n^{2(\alpha-1)}}+ o\left(\frac{1}{n^{2(\alpha-1)}}\right).$$

Theorems of comparison for positive series should then lead you to the conclusion: you need $2(\alpha-1)> 1$.

Hint. By using Taylor series expansions, as $n \to \infty$, we have $$\frac{\sinh \frac1n}{\sin \frac1n} = 1+\frac{1}{3 n^2}+O\left(\frac{1}{n^4}\right)$$ and $$(n^2+1)\log^\alpha \left(\frac{\sinh\frac{1}{n}}{\sin\frac{1}{n}}\right)=\frac{1}{3^a n^{2\alpha-2}}+O\left(\frac{1}{n^{2\alpha}}\right)$$ the given series is then convergent iff we have $2\alpha-2>1$.

• $$\frac{\sinh \frac1n}{\sin \frac1n} = 1+\frac{1}{3 n^2}+O\left(\frac{1}{n^4}\right)$$ how to get this expansions please? – Student Nov 13 '16 at 20:09
• @Kira Combining/composing that of $\sinh$, $\sin$, and $\frac{1}{1+x}$ when $x\to 0$. – Clement C. Nov 13 '16 at 20:16