calculate the sum of an infinite series

Question

Here is the series:

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

I don't know how to start at all. Thank you for your help.

Answer 1

Let $S=\sum_{k=0}^\infty\dfrac{x^{2k+1}}{2k+1}$

$$\dfrac{dS}{dx}=\sum_{k=0}^\infty(x^2)^k=\dfrac1{1-x^2}$$ for $|x^2|<1$

Integrate both sides to get $$S=\ln\dfrac{1+x}{1-x}+K$$

$x=0\implies0=\ln1+K\iff K=0$

Answer 2

$$\dfrac{(-1)^k}{2k+1}=-i\cdot\dfrac{i^{2k+1}}{2k+1}$$

Now for $-1\le x<1,$ $$\ln(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\dfrac{x^4}4+\cdots$$

$$\ln(1+x)-\ln(1-x)=?$$

Answer 3

I'm a student, so I may not have the mathematical rigour necessary, but I'll try to help.

It turns out that the Taylor Series for $\arctan(x)$ is this:

$$\arctan(x) = \sum_{k=0}^\infty (-1)^k\frac{(x^{2k+1})}{2k+1} = \frac{x}{1}-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...$$

So if x=1, you get:

$$\arctan(1) = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} = \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...$$

Which is exactly your series. Since $\arctan(1)=\frac{\pi}{4}$, your series converges to $\frac{\pi}{4}$. So

$$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1} = \frac{\pi}{4}$$

Some other things: How do I know your series converges? Well, because of Leibniz's rule for alternating series, which you'll find here.

How do I know that's the Taylor Series for $\arctan(x)$? If you take the derivative of $\arctan(x)$, which is $\frac{1}{1+x^2}$, and now you take its Taylor series and integrate it, you get the Taylor Series for $\arctan(x)$. Here's a Quora Post that explains how to obtain it:

(The links are spaced out because Quora won't let me otherwise)