The first part is straightforward, just plug in x = 1. For the second part, what is meant by compare? Are you asked to show which one converges faster per the number of terms?
Btw, the Wallis product for pi is
[Math]\displaystyle \prod_{n=1}^{\infty} \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2}[/tex]
The Leibnitz series is...
$\displaystyle \displaystyle \frac{\pi}{4} = (1-\frac{1}{3}) + (\frac{1}{5} - \frac{1}{7}) + ... + (\frac{1}{2n-1} - \frac{1}{2n+1}) + ...$ (1)
... so that after 2n iterations the last summed term is $\displaystyle \frac{1}{2n-1} - \frac{1}{2n+1} = \frac{2}{4n^{2}-1}$...
The Wallis product [in logarithmic version...] is...
$\displaystyle \displaystyle \ln \frac{\pi}{2} = \ln (1+\frac{1}{3}) + \ln (1 + \frac{1}{15}) + ... + \ln(1+\frac{1}{4n^{2}-1}) + ... $ (2)
... so that after 2n iterations the last summed term is $\displaystyle \ln (1+ \frac{1}{4 n^{2} -1}) \sim \frac{1}{4 n^{2}-1}$ ...
The two methods seem to be comparable in speed: both are very slow ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$