Proving that $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \log 2$ 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!
 A: 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.
A: 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
$$\lim_{n\to+\infty}\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}=f(1)=\ln(2).$$
A: Fom the Taylor expansion of the natural logarithm 
$$\ln(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}nx^n$$
With $x=1$, $\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln(2)$
