I have to determine the following sum:
\begin{equation} \sum\limits_{n=0}^{\infty}\frac{(-1)^n}{2n+1} \end{equation}
And i've tried lot's of ways to determine it, but i couldn't get a result so I figured I would look on WolframAlpha to see what the result has to be so i could maybe know what to do next by that. But apparently the result is $\frac{\pi}{4}$ and i don't know how I could possibly get to pi. Does anyone have any suggestions about how to tackle this?
Observe $$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$
$$\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}$$
Integrating both sides from $0$ to $1$ we get $$\arctan(x)|_0^1=\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}|_0^1$$ so $$\arctan(1)=\frac{\pi}{4}=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}$$
$$\frac{d(\tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$$ This is because, let $y=\tan^{-1}(x)$, so that $$x=\tan(y)\tag{1}$$ Differentiating both sides of $(1)$ yields $$1=\frac{1}{\cos^2(y)}\frac{dy}{dx}\\ 1=\frac{\cos^2(y)+\sin^2(y)}{\cos^2(y)}\frac{dy}{dx}\\\\1=\left(1+x^2\right)\frac{dy}{dx}$$ $$\boxed{\frac{d(\tan^{-1}(x))}{dx}=\frac{1}{1+x^2}}\tag{2}$$
Now, it is known that $$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$$ for $|x|<1$, replacing $x$ with $-x^2$ we get $$\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^nx^{2n}\tag{3}$$ Combining $(2)$ and $(3)$ we see that $$\frac{d(\tan^{-1}(x))}{dx}=\sum_{n=0}^{\infty}(-1)^nx^{2n}\tag{4}$$ Taking the integral from $0$ to $1$ of $(4)$ yields $$\int_0^1\frac{d(\tan^{-1}(x))}{dx}=\tan^{-1}(1)-\tan^{-1}(0)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\\\boxed{\frac{\pi}{4}=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}}$$
As others have shown, $$\sum_{n\geq0}\frac{(-1)^n}{2n+1}=\int_0^1\frac{\mathrm dx}{1+x^2}$$ but we can generalize this.
$$\frac1{1+x}=\sum_{n\geq0}(-1)^nx^n$$ $$\frac1{1+x^a}=\sum_{n\geq0}(-1)^nx^{an}$$ $$\int_0^1\frac{\mathrm dx}{1+x^a}=\sum_{n\geq0}\frac{(-1)^n}{an+1}$$ Also $$\int_0^b\frac{\mathrm dx}{1+x^a}=\sum_{n\geq0}\frac{(-1)^nb^{an+1}}{an+1}$$
Hint: $\sum_{k=0}^n{\frac{(-1)^k}{2k+1}} = \int_0^1{\frac{1}{1+x^2}} - \int_0^1{\frac{(-x^2)^{n+1}}{1+x^2}}$.
The method I give can be applied to a number of sums, as I have been discovering.
We have
\begin{align*} \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} & = \sum_{n=1}^\infty \dfrac{\sin \dfrac{\pi n}{2}}{n} \nonumber \\ & = \sum_{n=1}^\infty \dfrac{\sin \dfrac{\pi n}{2}}{n} \int_0^\infty e^{-y} dy \nonumber \\ & = \sum_{n=1}^\infty \sin \dfrac{\pi n}{2} \int_0^\infty e^{-nx} dx \nonumber \\ & = \frac{1}{2i} \sum_{n=1}^\infty \int_0^\infty (e^{-nx + \frac{i\pi n}{2}} - e^{-nx + \frac{-i\pi n}{2}}) dx \nonumber \\ & = \frac{1}{2i} \int_0^\infty \left( \dfrac{1}{1-e^{-x + \frac{i\pi}{2}}} - \dfrac{1}{1-e^{-x + \frac{-i\pi}{2}}} \right) dx \nonumber \\ & = \int_0^\infty \dfrac{1} {e^x + e^{-x}} dx \end{align*}
You can use the fact that $\frac{d}{dx} \tan^{-1} (\sinh x) = \frac{1}{\cosh x}$ to obtain:
\begin{align*} \frac{1}{2} \int_0^\infty \dfrac{2}{e^x + e^{-x}} dx = \frac{1}{2} [\tan^{-1} (\sinh x)]_0^\infty = \frac{\pi}{4} . \end{align*}
As an alternative, using that result, we have
$$\sum\limits_{n=0}^{N}\frac{(-1)^n}{2n+1}=\sum_{n=0}^{N}(-1)^n\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)+\sum_{n=0}^{N+1}\frac{(-1)^n}{2n}\to \frac{\pi}{4}-\frac{\log 2}{2}+\frac{\log 2}{2}=\frac{\pi}{4}$$
