Prove (Leibniz' series): $|s_n - \frac \pi 4| \le \frac 1 {2n+1}, \forall n \in \mathbb N$ where $s_n = \sum^{n-1}_{j=0} \frac {(-1)^j} {2j+1} = 1 - \frac 1 3 + \frac 1 5$ ...
To prove the result $\sum^{\infty}_{j=0} \frac {(-1)^j} {2j+1} = \frac \pi 4$ known as Leibniz' series, I've already proved:
(1) $\int^1_0 \frac 1 {1+x^2} dx = s_n + (-1)^n \int^1_0 \frac {x^{2n}} {1+x^2} dx, \forall n \in \mathbb N$
(2) $0 \le \int^1_0 \frac {x^{2n}} {1+x^2} dx \le \frac 1 {2n+1}, \forall n \in \mathbb N$
I'm now asked to prove $\forall n \in \mathbb N : |s_n - \frac \pi 4| \le \frac 1 {2n+1}$, which can be used to prove that $\lim_{n \rightarrow \infty} \sum^{n-1}_{j=0} \frac {(-1)^j} {2j+1} = \frac \pi 4$.
To be honest I've no idea how to relate $s_n$ to $\frac \pi 4$ - There must be some relation in terms of (1) and (2) ?
In general, if $a_n>0$, $a_n\to 0$ and $a_n$ decreasing then the series $$ \sum_{n=1}^\infty (-1)^{n-1} a_n, $$ converges (not necessarily absolutely), say to $s$, and $$ s_{2n-1}<s_{2n+1}<s<s_{2n+2}<s_{2n}, $$ where $s_n=\sum_{k=1}^n (-1)^{k-1}a_k$. In particular, $$ s_n-s\quad\text{and}\quad s_{n+1}-s, $$ have different signs. This implies that $$ \lvert s-s_{n-1}\rvert<a_n. $$ In our case $$ \frac{\pi}{4}-s_n=\int_0^1\frac{dx}{1+x^2}-s_n=(-1)^n\int_0^1\frac{x^{2n}\,dx}{1+x^2} $$ and since $\int_0^1\frac{x^{2n}\,dx}{1+x^2}\le\frac{1}{2n+1}$, then
$$ \left\lvert \tfrac{\pi}{4}-s_{n}\right\rvert<\frac{1}{2n+1}, $$ which implies that $$ \lim_{n\to\infty} s_n=\frac{\pi}{4}. $$