Prove that $\sum_{k=0}^{\infty} (k-1)/2^k = 0$ How to prove that this series converges, and that the limit is 0 ?

You don't need calculus at all.
The sum is equal to $-1+ 0 + \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots$
In essence, we want to show that $\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots = 1$
We can decompose this into an infinite number of infinite geometric series as follows:
$$S_1 = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{1}{2}$$
$$S_2 = \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots = \frac{\frac{1}{8}}{1 - \frac{1}{2}} = \frac{1}{4}$$
$$S_3 = \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \dots = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{1}{8}$$
And so forth.
Furthermore, notice that $$S = S_1 + S_2 + S_3 + \dots = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$$ is itself geometric! In addition, it's sum is $1$! Thus, we have just shown that $\displaystyle \sum_{k = 0}^{\infty} \frac{(k-1)}{2^k} = 0$
