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 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 $$\int_0^1 \frac {dx}{1+x}=\ln(2)$$