What is the series $\sum_{i=n}^{\infty}\frac{i}{2^i}$ When computing the expected value for a random variable I reached the following series:
$$\sum_{i=n}^{\infty}\frac{i}{2^i}$$
I am confident it is convergent, but have no idea how to compute it.
 A: There are a few standard tricks that one can use.  For example:
Let $\displaystyle f(x) = \sum_{i=n}^\infty i x^{i-1} = \frac{d}{dx} \sum_{i=n}^\infty x^i = \frac{d}{dx} \frac{x^n}{1-x} = \frac{nx^{n-1}(1-x) + x^n }{(1-x)^2}$.
So $\displaystyle \sum_{i=n}^\infty \frac{i}{2^i} = \frac{1}{2} \sum_{i=n}^\infty i \left(\frac{1}{2}\right)^{i-1} = \frac{1}{2} f\left(\frac{1}{2}\right) = \frac{1}{2} \frac{n(1/2)^{n-1} (1/2)+(1/2)^n}{1/4} = \frac{n+1}{2^{n-1}}$.
A: $$ \sum_{i = n}^{\infty} \frac{i}{2^i}$$
$$\frac{i}{2^i} < 1, \space \forall i \ge 0$$
$$ \lim_{i\to\infty}{\frac{i}{2^i}} \to0$$
And apply ratio test of Convergence 
$$ \lim_{n\to\infty}\frac{U_{n+1}}{U_n} = \frac{n\times{2^{n+1}}}{(n+1)\times{2^n}} = \frac{n}{(n+1)}\times 2 = 2 < M$$
Here $M$ is finite positive no. Hence the given series will converge.
A: We can sum your series by interpreting it as the expectation of a random variable, and then computing that expectation in another way.
Suppose a fair coin is tossed repeatedly until it comes up heads. Let $X$ be the number of tosses, up to and including the first head. Let $I=X[X\ge n]$, i.e., $I=X$ if $X\ge n$, and $I=0$ otherwise. It's easy to see that$$E[I]=\sum_{i=n}^{\infty}\frac i{2^i}$$which is the series you asked about.
Let $Y_i$ be the indicator variable whose value is $1$ if $X\ge i$ and $0$ otherwise. Clearly $E[Y_i]=\dfrac1{2^{i-1}}$, the probability of no heads in the first $i-1$ tosses. Moreover, it's easy to see that$$I=nY_n+\sum_{i=n+1}^{\infty}Y_i,$$whence$$E[I]=nE[Y_n]+\sum_{i=n+1}^{\infty}E[Y_i]=\frac n{2^{n-1}}+\sum_{i=n+1}^{\infty}\frac1{2^{i-1}}=\frac n{2^{n-1}}+\frac1{2^{n-1}}=\frac{n+1}{2^{n-1}}.$$
Edit. $n$ is a fixed natural number; $X$ is a certain discrete random variable whose values are nonnegative integers. 
For each $i\in\mathbb N$, the random variable $Y_i$ is a function of $X$, namely, $Y_i$ takes the value $0$ when $X\lt i$, the value $1$ when $X\ge i$. In terms of the discrete form of the Heaviside step function, $Y_i=H[X-i]$; in the Iverson bracket notation, $Y_i=[X\ge i]$.
We also defined a random variable $I=XY_n$; when $X\lt n$ we have $I=X\cdot0=0$, when $X\ge n$ we have $I=X\cdot1=X$.
Now we verify the identity$$I=nY_n+\sum_{i=n+1}^{\infty}Y_i.$$Let $k$ be the observed value of the random variable $X$. By our definition of $Y_i$, we have $Y_i=1$ for all $i\le k$, and $Y_i=0$ for all $i\gt k$.
If $k\lt n$, then all terms on both sides of the identity are zero.
If $k=n$, then the identity reduces to $$n=n\cdot1+\sum_{i=n+1}^{\infty}0.$$
If $k=n+r$ for some $r\in\mathbb N$, then the identity reduces to $$n+r=n\cdot1+\sum_{i=n+1}^{n+r}1+\sum_{i=n+r+1}^{\infty}0.$$
A: Assume we know that (see this link)
$\tag 1 \displaystyle{\sum_{i=0}^{\infty}\frac{i}{2^i} = 2}$
For any integer fixed integer $n \ge 0$ we have a statement $S(n)$ defined by
$\tag 2 S(n): \quad \displaystyle{\sum_{i=n}^{\infty}\frac{i}{2^i} = \frac{n+1}{2^{n-1}}}$
We will prove that
$\tag 3 k \ge 0 \text{ implies } S(k)$
Using $\text{(1)}$ and the fact that
$\quad 2 = \frac{0+1}{2^{0-1}}$
it follows that $S(0)$ is true.
Assume $S(n)$ is true. This is equivalent to writing
$\tag 4  \displaystyle{\sum_{i=n}^{\infty}\frac{i}{2^i} - \frac{n}{2^n} = \frac{n+1}{2^{n-1}} - \frac{n}{2^n} }$
The LHS of $\text{(4)}$ is equal to $\displaystyle{\sum_{i=n+1}^{\infty}\frac{i}{2^i}}$.
As for the RHS, applying algebra we can write
$\quad \frac{n+1}{2^{n-1}} - \frac{n}{2^n} = \frac{(n+1)+1}{2^{(n+1)-1}}$
We have shown that $S(n+1)$ must also be true.
By induction we conclude the veracity of $\text{(3)}$.
