summation - Finite Sum $sum_{i=1}^nfrac i {2^i}$

I'm trying to find the sum of :

$$\sum_{i=1}^n\frac i {2^i}$$

I've tried to run $i$ from $1$ to $∞$ , and found that the sum is $2$ , i.e :

$$\sum_{i=1}^\infty\frac i {2^i} =2$$

since :

$$(1/2 + 1/4 + 1/8 + \cdots) + (1/4 + 1/8 + 1/16 +\cdots) + (1/8 + 1/16 + \cdots) +\cdots = 1+ 1/2 + 1/4 + 1/8 + \cdots = 2 $$

But when I run $i$ to $n$ , it's a little bit different , can anyone please explain ?

Regards

Answer

For the sake of generality, and more importantly to make typing easier, we use $t$ instead of $1/2$.

We want to find the sum

$$S(t)=t+2t^2+3t^3+4t^4+ \cdots +(n-1)t^{n-1}+nt^n.$$
Multiplying both sides by $t$, we get
$$tS(t)=t^2+2t^3+3t^4 +4t^5+ \cdots +(n-1)t^{n}+nt^{n+1}.$$
Subtract, and rearrange a bit. We get
$$(1-t)S(t)=(t+t^2+t^3+ +\cdots +t^n)-nt^{n+1}.\tag{$\ast$}$$
Recall that for $t\ne 1$, we have $t+t^2+t^3+ +\cdots +t^n=t\frac{1-t^n}{1-t}$.
If we do not recall the sum of a finite geometric series, we can find it by a trick similar to (but simpler) than the trick that got us to $(\ast)$.

Substitute, and solve for $S(t)$. (The method breaks down when $t=1$ because of a division by $0$ issue. But $t=1$ is easy to handle separately.)

Remark: Now that we have obtained an expression for $\sum_{k=1}^n kt^k$, we can use this expression, and the same basic trick, to find $\sum_{k=1}^n k^2t^k$, and then $\sum_{k=1}^n k^3t^k$. Things get rapidly more unpleasant, and to get much further one needs to introduce new ideas.