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

Question

As stated in the title, I am trying to prove that $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \log2$, I think I need to use the Geometric Series like this:

$$1+u+u^2+...+u^n = \frac{1-u^{n+1}}{1-u}$$

Then setting $u = -t$, rewrite it like so:

$$1-t+t^2-...+(-1)^n t^n + \frac{(-1)^{n+1}t^{n+1}}{1+t}= \frac{1}{1+t}$$

But now I am not sure where to go next in order to get to my conclusion, any guidance and explanation would be appreciated, thanks!

Answer 1

Take the integral from $0$ to $1$ on both sides of the equation, yielding

$$1 - \frac{1}{2} + \frac{1}{3} -\cdots + (-1)^n\frac{1}{n+1} + (-1)^{n+1}\int_0^1 \frac{t^{n+1}}{1+t}\, dt = \log 2$$

We want to show

$$\lim_{n\to\infty} (-1)^{n+1}\int_0^1 \frac{t^{n+1}}{1+t}\, dt = 0\tag{*}$$

To do so, we use the inequality $1 + t \ge 1$ for $t \ge 0$ to obtain

$$\left\lvert (-1)^{n-1} \int_0^1 \frac{t^{n+1}}{1+t}\, dt\right\rvert = \int_0^1 \frac{t^{n+1}}{1+t}\, dt \le \int_0^1 t^{n+1}\, dt = \frac{1}{n+2}$$

Since $\lim_{n\to \infty} \frac{1}{n+2} = 0$, (*) holds.

Answer 2

hint

Let $f(x)=\ln(1+x)$. $f$ is $C^{\infty}$ near $0$. thus for $n>0, \exists c\in(0,1)\;:\;$

$$f(1)=f(0)+\sum_{k=1}^n\frac{1}{k!}f^{(k)}(0)+\frac{1}{(n+1)!}f^{(n+1)}(c)$$

but $f^{(n+1)}(c)=(-1)^{n}\frac{n!}{(1+c)^{n+1}}$

and $0<\frac{1}{1+c}<1$ , so

$$\lim_{n\to+\infty}\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}=f(1)=\ln(2).$$

Answer 3

Fom the Taylor expansion of the natural logarithm

$$\ln(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}nx^n$$

With $x=1$, $\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln(2)$