Convergence of $\sum_{n = 1}^{\infty} \frac{(-1)^n}{n}$ I was messing around in Mathematica with infinite sums until I tried taking the sum of the following:
$$\sum_{n = 1}^{\infty} \frac{(-1)^n}{n}$$
Mathematica spat out -Log[2]. Can somebody give me proof explaining this answer?
 A: One may start with the standard geometric series identity,
$$
1-x+x^2-\cdots+(-1)^Nx^{N}=\frac{1+(-1)^Nx^{N+1}}{1+x},\quad x \neq-1,
$$ giving
$$
\int_0^1\left(1-x+x^2-\cdots+(-1)^Nx^N\right)dx=\int_0^1\frac{1+(-1)^Nx^{N+1}}{1+x}\:dx
$$ or
$$
\sum_{n = 1}^{N+1} \frac{(-1)^{n-1}}{n}=\int_0^1\frac1{1+x}\:dx+\int_0^1\frac{(-1)^Nx^{N+1}}{1+x}\:dx=\ln2+\int_0^1\frac{(-1)^Nx^{N+1}}{1+x}\:dx
$$ then, letting $N \to \infty$, one may use
$$
\left|\int_0^1\frac{(-1)^Nx^{N+1}}{1+x}\:dx\right|\le \int_0^1x^{N+1}dx=\frac1{N+2} \to 0,
$$ which gives

$$
\sum_{n = 1}^\infty \frac{(-1)^{n}}{n}=-\ln 2.
$$

A: A formal argument (ignoring convergence questions)
Start with the Geometric series $$\frac 1{1-x}=1+x+x^2+\cdots$$
Integrate to obtain $$-\ln(1-x)=x+\frac {x^2}2+\frac {x^3}3+\cdots$$
Now evaluate at $x=-1$ to get your result.
To be more rigorous, note that (inductively) it is easy to prove 
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
And the right hand can then be rewritten as $$\frac{1}{n} \left[ \frac{1}{1+\frac{1}{n}}+ \frac{1}{1+\frac{2}{n}}+\cdots+\frac{1}{1+\frac{n}{n}} \right]$$
Which is the standard Riemann sum approximation to $$\int_0^1 \frac {dx}{1+x}=\ln(2)$$
