1. ## series convergence

study the convergence of the following series indicating the radius of convergence and the interval of convergence

i) $
\sum\limits_{n = 1}^\infty {sen\left( {\pi \left( {n + \frac{1}{n}} \right)} \right)} x^n
$

ii) $
\sum\limits_{n = 1}^\infty {\ln \left( {1 + \frac{1}{{n^2 }}} \right)} x^n

$

please i need help with this to understand how to do it D:

2. When you've written "sen", i believe you meant "sin". If that is so, you can try working on the function. You should check out the formula

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After that, you gotta remember what the criteria for convergence of sums like [LaTeX ERROR: Convert failed] are. Also, what formulas involving [LaTeX ERROR: Convert failed] and limits do you have?

3. $\displaystyle{\ln\left(1+\frac{1}{n^2}\right) = \ln\left(\frac{n^2+1}{n^2}\right) = \ln\left(n^2+1\right)-2\ln n}$

Let's use the ratio test: $\displaystyle{\lim_{n\to\infty}\left|\frac{a_{n+1} }{a_n}\right| = |x|\cdot\lim_{n\to\infty}\left|\frac{\ln\left((n+1 )^2+1\right)-2\ln(n+1)}{\ln\left(n^2+1\right)-2\ln n}\right|}}$

(Applying L'Hopital's rule) $\displaystyle{= |x|\cdot\lim_{n\to\infty}\left|\frac{\frac{2(n+1)} {(n+1)^2+1}-\frac{2}{n+1}}{\frac{2n}{n^2+1}-\frac2n}\right|} = |x|\cdot\lim_{n\to\infty}\left|\frac{\frac{2}{(n+1 )(n^2+2n+2)}}{\frac{2}{n(n^2+1)}}\right|}}$

$\displaystyle{=|x|\cdot\lim_{n\to\infty}\left|\fra c{n^3+n}{n^3+3n^2+4n+2}\right|}} = |x|\leq1$

That just leaves us to consider the end points: $x=\pm1$.

When $x=-1$ we apply the alternating series test to see it converges. I leave the case of $x=1$ for you.

So in the end we get the interval of convergence is $[-1,1]$.

4. Considering the identity...

$\displaystyle \sin \pi\ (n+\frac{1}{n}) = (-1)^{n} \sin \frac{\pi}{n}$ (1)

... and that for 'large' n is $\sin \frac{\pi}{n} \sim \frac{\pi}{n}$, the two series...

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

$\displaystyle \sum_{n=1}^{\infty} (-1)^{n}\ \frac{x^{n}}{n}$ (2)

... have the same radious of convergence...

For 'large n' is $\ln (1+ \frac{1}{n^2}) \sim \frac{1}{n^{2}}$, so that the two series...

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

$\displaystyle \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}}$ (3)

... have the same radious of convergence...

Kind regards

$\chi$ $\sigma$