# Compute the sum of the infinite series $\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$

Is it possible to directly show $$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}=\frac{\pi}4$$ using partial sums?

I think I was able to do something else to get this result as follows, which I posted separately as my own answer to this question.

However, I was unable to compute the sum using $$n$$-th partial sums, but I would like to ask if someone knows how to do that here. Writing out the first four terms $$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}=1-\frac 13+\frac 15-\frac 17+\cdots$$ the series is not telescoping (wishful thinking...), so that would not help us here. It also seems that we cannot use partial fractions for this series. But I hope what I posted as an answer is not the only way to compute the sum.

Starting from the geometric sum formula $$\frac 1{1+x^2}=\sum_{k=0}^\infty (-1)^kx^{2k}$$ for all $$x \in (-1,1)$$, we integrate both sides to get $$\tan^{-1}x=\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}x^{2k+1}.$$ We send $$x \to 1^-$$ of both sides to conclude $$\frac{\pi}4=\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}.$$

• Note: You need Abel's theorem when you take the limit. – Botond Sep 12 at 19:05

For $$-1\le x<1,$$ $$\ln(1-x)=-\sum_{r=1}^\infty\dfrac{x^r}r$$

So, $$\ln\dfrac{1+x}{1-x}=\ln(1+x)-\ln(1-x)=?$$

If $$S=\sum_{k=0}^\infty\dfrac{(-1)^k}{2k+1},$$

$$2iS=\ln\dfrac{1+i}{1-i}=\ln(i)$$ whose principal value is $$\dfrac{i\pi}2$$

Considering that $$\frac{1}{2k+1}=\int_0^1{x^{2k}dx},$$ we have that $$\sum_{k=0}^n{\frac{(-1)^k}{2k+1}}=\int_0^1{\frac{dx}{1+x^2}}-\int_0^1{\frac{(-x^2)^{n+1}}{1+x^2}dx}=\frac{\pi}{4}+(-1)^{n}\int_0^1{\frac{x^{2n+2}}{1+x^2}dx}.$$ However $$0\leqslant\int_0^1{\frac{x^{2n+2}}{1+x^2}dx}\leqslant\int_0^1{x^{2n+2}dx}=\frac{1}{2n+3}\underset{n\rightarrow +\infty}{\longrightarrow}0$$ since $$\forall x\in[0,1],\,\frac{1}{1+x^2}\leqslant 1$$. Letting $$n\rightarrow +\infty$$ gives the result.