I can and I have succesfully completed question 6 out of 17!!

I would never have seen the substitution n=2k+1. It is a little ingenious.

Right, we haven't been shown yet how to determine what it converges to. I can show you it converges though (which is a little easy, I know):

$\displaystyle \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$=$\displaystyle -\sum_{k=0}^\infty \frac{(-1)^{k+1}}{2k+1}$

and by the alternating series test this converges since $\displaystyle \frac{1}{2k+1} \rightarrow 0 \ as \ k \rightarrow 0$

so what does the series converge to?