I am not sure how to find the sum of this series
$$\sum_{n=0}^{\infty}{(-1)^n\cdot(2n+1)^{-1}}$$
I know it converges due to the alternating series test because the function is decreasing over its domain and the limit as $n$ approachs infinite is zero. However, I don't know what method to use to compute the finite value it converges to.
Also I know the answer is $\pi/4$ because Wolframalpha but I want to know how. Thanks.
Best Answer
This can be shown from the Taylor series
$$ \arctan x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} $$
Letting $x=1$
$$ \sum_{n=0}^\infty (-1)^n \frac{1}{2n+1} = \arctan 1 = \frac{\pi}{4} $$
For a proof of this, infer from the geometric series $$ \frac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n} $$
Then integrate both sides