Prove that $\sum_{k=0}^{\infty} (k-1)/2^k = 0$ How to prove that this series converges, and that the limit is 0 ?
 A: Re-write the sum like so:
$$\sum_{k=0}^{\infty} \frac{k-1}{2^k} = -1 + \frac{1}{2} \sum_{k=0}^{\infty} \frac{k}{2^k}$$
To evaluate the right sum, for $|x|<1$, consider: 
$$\begin{align*}f(x) = \sum_{k=0}^{\infty} x^k = \frac{1}{1-x} \implies f'(x) = \sum_{k=0}^{\infty} k x^{k-1}= \frac{1}{(1-x)^2}\implies xf'(x) & = \sum_{k=0}^{\infty} k x^k \\ & = \frac{x}{(1-x)^2}\end{align*} $$
Substitute $x=\tfrac{1}{2}$.
EDIT: The series can be shown to converge by using the ratio test.
A: $$\sum_{r=0}^n(r-1)x^r=x^2\sum_{r=0}^n\frac{d(x^{r-1})}{dx}$$
Again, $$\sum_{r=0}^n\frac{d(x^{r-1})}{dx}=\frac{d\left[\sum_{r=0}^n x^{r-1}\right]}{dx}$$
Now now summation formula and then set $n\to\infty$ for $x=\frac12$ and $|x|<1$
A: 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 $\displaystyle \sum_{k = 0}^{\infty} \frac{(k-1)}{2^k} = 0$
A: $$
\sum_{k=0}^\infty \frac{k-1}{2^k} = -1 + \frac{1}{2}\sum_{k=1}^\infty \frac{k-1}{2^{k-1}}.
$$
If we call $n=k-1$, then
$$
\sum_{k=0}^\infty \frac{k-1}{2^k} = -1 + \frac{1}{2}\sum_{n=0}^\infty \frac{n}{2^n} = -1 + \frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n}+ \frac{1}{2}\sum_{n=0}^\infty \frac{{n-1}}{2^n}.
$$
Therefore
$$
\frac{1}{2}\sum_{k=0}^\infty \frac{{k-1}}{2^k} = -1 + \frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n} = 0.
$$
