Prove $\frac{\pi}{4} = \sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}$ using Dominated or Monotone Convergence Is there a way to prove that $$\frac{\pi}{4} = \sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}$$ via the Dominated or Monotone Convergence Theorem?
 A: Note that
$$\frac{1}{2n+1}=\int_0^1 t^{2n}dt.$$
So 
$$\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}=\sum_{n=0}^\infty\int_0^1 (-1)^nt^{2n}dt=\sum_{n=0}^\infty\int_0^1 (-t^2)^ndt=\int_0^1\sum_{n=0}^\infty(-t^2)^ndt=\int_0^1\frac{dt}{1+t^2}=\frac\pi2$$
To justify the interchange $\sum\int=\int\sum$ we use the dominated convergence theorem:
$$\int_0^1\sum_{n=0}^\infty(-t^2)^ndt=\lim_{N\to\infty}\int_0^1\sum_{n=0}^N(-t^2)^ndt=\lim_{N\to\infty}\int_0^1f_N(t)dt$$
and 
$$|f_N(t)|=\left|\frac{1-t^{2N+2}}{1+t^2}\right|\le\frac2{1+t^2}\in L^1(0,1)$$
A: Yes. Consider the power series for $\tan^{-1}$. You will see that if you plug in $x=1$ into this series and equate it with $\tan^{-1}(1)$, you get what you want. 
However, you only know that $\tan^{-1}$ equals this series inside the radius of convergence ($|x|<1$). So you should use Abel's theorem (see the simplified version in the 'Remarks' section) to show that the limit as $x\rightarrow 1$ from below is $\tan^{-1}(1)$, then use your convergence theorems to compute this limit (which is the series in your question).
A: You could also start recongnizing that $$S_1=\sum_{n = 0}^{\infty} (-1)^n {t^{2n}}$$ is simply the series expansion of $\frac{1}{1+t^2}$ and then that
$$S_2=\sum_{n = 0}^{\infty}\frac{(-1)^{n}t^{2n+1}}{2n + 1}$$ is just the antiderivative of $S_1$. So, $S_2=\tan ^{-1}(t)$. Now, make $t=1$ in this last expression and get $S_2=\frac{\pi}{4}$
