Hello
Find the sum of the following series :
1- $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n {\pi}^n}{9^n (2n)!}$
2- $\displaystyle \sum_{n=2} ln(1-\frac{1}{n^2})$
hmmm I do not have any ideas
1-
$\displaystyle \color{blue}{\pi}^n=(\sqrt{\pi})^{2n}$
$\displaystyle \color{blue}9^n=3^{2n}$
Now, think about a well-known maclaurin series.
2-
Hint:
$\displaystyle \color{blue}ln(1-\frac{1}{n^2})=ln(\frac{n^2-1}{n^2})=ln(\frac{(n-1)(n+1)}{(n)(n)})=ln(\frac{n-1}{n})+ln(\frac{n+1}{n})$
The 'infinite sum'...
$\displaystyle \sum_{n=2}^{\infty} \ln (1-\frac{1}{n^{2}})$ (1)
... is , in my opinion of course, 'very nice'. If we remember the 'infinite product'...
$\displaystyle sinc (x)= \frac{sin \pi x}{\pi x} = \prod_{n=1}^{\infty} (1-\frac{x^{2}}{n^{2}})$ (2)
... it is easy to derive that is...
$\displaystyle \prod_{n=2}^{\infty} (1-\frac{1}{n^{2}}) = \lim_{x \rightarrow 1} \frac{sinc (x)}{1-x^{2}} = \frac{1}{2}$ (3)
... so that...
$\displaystyle \sum_{n=2}^{\infty} \ln (1-\frac{1}{n^{2}})= -\ln 2 $ (4)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$