I'm having a bit of a dilemma right now and I thought I'd ask here. Pretty much, I have a list of known Maclaurin series that I can use on my exam and they greatly help me in solving series and their sums. However, I've encountered a bit different problem today. I have a series that in almost every aspect matches the one in the Maclaurin series list, with one difference: the lower bound of the series is $n=1$ as opposed to $n=0$ (the way it is written in the list). I'll give you the exact task I have as well:
The problem I need to solve (find the sum):
$$\sum_{n=1}^\infty (-1)^n \frac{\pi^{2n+1}}{4^{2n+1} (2n+1)!}$$
The Maclaurin series that I've found:
$$\sin x=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$
I've tried doing it as if the lower bound was the same, and I get $\frac{1}{4}\sin(\pi)\sum_{n=0}^\infty \frac{1}{4^{2n}}$
Can I do this? Does the lower bound matter in this situation?
Of course the lower bound matters, but in this case (as noted already in comments) you just have to subtract the first term in order to start the series one term later.
A bigger conceptual problem is that you seem to think that
$$ \sum_{n=n_0}^\infty f(n)g(n) \stackrel?= \left(\sum_{n=n_0}^\infty f(n)\right)\left(\sum_{n=n_0}^\infty g(n)\right). $$
This is not true in general.