$\sum_{n=0}^{\infty} (-1)^n (\frac{31}{32})^{2n+1} (\frac{1}{2n+1})$
Hey math stack exchange! I've been trying to figure out how to calculate the exact summation of these series for awhile now. I looked at this post: How to find the sum of an alternating series? to get some clues, but I'm still a bit lost because my formula has the term $(\frac{1}{2n+1})$ in it.
Would appreciate some help! Thanks!
$$S = \sum_{n=0}^{\infty}{(-1)^{n}\biggl(\frac{31}{32}\biggr)^{2n+1}\biggl(\frac{1}{2n+1}\biggr)}$$ $$= \sum_{n=0}^{\infty}{(-1)^{n}\Biggl(\int_{0}^{\frac{31}{32}}x^{2n}dx\Biggr)}$$ $$= \int_{0}^{\frac{31}{32}}\Biggl(\sum_{n=0}^{\infty}(-x^{2})^{n}\Biggr)dx$$ $$= \int_{0}^{\frac{31}{32}}{\Biggl(\frac{1}{1+x^{2}}\Biggr)}dx$$ $$= \arctan\Biggl({\frac{31}{32}}\Biggr)$$
We first look at the power series $f(x) = \sum_{n = 0}^\infty \frac{(-1)^n}{2n + 1}x^{2n + 1}$. It is clear that the question is asking for the value of $f(\frac{31}{32})$.
We now take derivative of $f(x)$ and get $$f'(x) = \sum_{n = 0}^\infty (-1)^nx^{2n} = \frac 1{1 + x^2}.$$ Special attention should be paid to the convergence problem, but I'll omit details here.
Now we integrate and get $f(x) = \arctan(x)$, noting that $f(0) = 0$.
