Prove that $\sum_{k=0}^{\infty} (k-1)/2^k = 0$

Question

How to prove that this series converges, and that the limit is 0 ?

Answer 1

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.

Answer 2

$$\sum_{r=0}^n(r-1)x^r=x^2\sum_{r=0}^n\frac{d(x^{r-1})}{dx}$$

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 $|x|<1$

Answer 3

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 $\displaystyle \sum_{k = 0}^{\infty} \frac{(k-1)}{2^k} = 0$

Answer 4

$$ \sum_{k=0}^\infty \frac{k-1}{2^k} = -1 + \frac{1}{2}\sum_{k=1}^\infty \frac{k-1}{2^{k-1}}. $$ 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 $$ \frac{1}{2}\sum_{k=0}^\infty \frac{{k-1}}{2^k} = -1 + \frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n} = 0. $$