Process to get to the $\frac{\pi}{4}$ integral formula

Question

What is the process in between the sum:

$\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$

to get to the integral: $\int_{0}^{1} \frac{1}{1+x^2}dx$

Answer 1

Note that $$\frac{1}{2k+1}=\int_{0}^{1}x^{2k}\,dx$$ and hence $$\int_{0}^{1}\frac{dx}{1+x^{2}}-\sum_{k=0}^{n}(-1)^{k}\frac{1}{2k+1}=\int_{0}^{1}\left(\frac{1}{1+x^{2}}-\sum_{k=0}^{n}(-1)^{k}x^{2k}\right)\,dx$$ The integral on the right can be expressed (using sum of a geometric progression) as $$(-1)^{n+1}\int_{0}^{1}\frac{x^{2n+2}}{1+x^{2}}\,dx=(-1)^{n+1}I_{n}\text{(say)}$$ and clearly $$0\leq I_{n} \leq \int_{0}^{1}x^{2n+2}\,dx=\frac{1}{2n+3}$$ By Squeeze Theorem $I_{n}\to 0$ and hence $(-1)^{n+1}I_{n}\to 0$ as $n\to\infty$. It follows now that $$\int_{0}^{1}\frac{dx}{1+x^{2}}=\sum_{k=0}^{\infty}(-1)^{k}\frac{1}{2k+1}$$ Note that the analysis of $I_{n} $ is easy but still necessary.

Answer 2

hint: $\dfrac{1}{2n+1} = \displaystyle \int_{0}^1 x^{2n}dx$. Can you take it from here?