Verification of this summation

Question

How do I check or evaluate this summation

$$\sum_{k\ge 0} \left(\frac12\right)^{k+1}k=1$$

Answer 1

Hint

Consider $$\sum_{k=0}^\infty x^{k+1}k=x^2\sum_{k=0}^\infty x^{k-1}k=x^2 \frac{d}{dx}\Big(\sum_{k=0}^\infty x^k\Big)=x^2\frac{d}{dx}\Big(\frac{1}{1-x}\Big)=\frac{x^2}{(1-x)^2}$$ Now replace $x$ by $\frac{1}{2}$ and you are done

Answer 2

For $|x| < 1$ we have: $\displaystyle \sum_{k=0}^\infty kx^{k-1} = \left(\dfrac{1}{1-x}\right)' = \dfrac{1}{(x-1)^2}$. Thus: $\displaystyle \sum_{k=0}^\infty kx^{k+1} = \dfrac{x^2}{(x-1)^2}$. Put $x = 1/2$ in the equation to get the answer: $1$

Answer 3

hints:

$$|x|<1\implies\frac1{1-x}=\sum_{n=0}^\infty x^n\implies\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}$$

What then happens with $\;x=\frac12\;$ and...?

Answer 4

Multiply by $2$. Note that the first term is $0$, so we want to show that $$\sum_1^\infty k\left(\frac{1}{2}\right)^k=2.\tag{1}$$

Define random variable $X$ as follows. Imagine tossing a fair coin until we get a head. Let $X$ be the number of tosses. Then for $k\ge 1$ we have $\Pr(X=k)=\frac{1}{2^k}$. It follows that the sum (1) is $E(X)$.

Let $e$ be the expectation of $X$. Condition on the first toss. If the first toss is a head, then $X=1$, so the conditional expectation of $X$, given that we get a head on the first toss, is $1$.

If the first toss is a tail, then we have used up one toss, and the expected number of additional tosses until we get a head is $e$. It follows that $$e=\frac{1}{2}\cdot 1+\frac{1}{2}\cdot(1+e).$$ Solve this linear equation for $e$. We get $e=2$.