Finite Sum $\sum_{i=1}^n\frac i {2^i}$ 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
 A: Let $r=1/2$.
Write all the terms being added as
$$ \left.
\matrix{
r& \phantom{2}r^2& \phantom{2}r^3& \phantom{2}r^4&\cdots&\phantom{n}r^n \cr
0 & \phantom{2}r^2& \phantom{2}r^3& \phantom{2}r^4&\cdots& \phantom{2}r^n \cr
0&0& \phantom{2}r^3& \phantom{2}r^4&\cdots& \phantom{2}r^n\cr
& \vdots&&&&\cr
0&0&0&0&\cdots& \phantom{2} r^n\cr
}\ \ \right\}  n-\text{rows}
$$
$$ 
\overline{\matrix{ r&2r^2&3r^3&4r^4&\cdots&nr^n}\phantom{dfgfsdfsfs}}
$$
Using the formula for the sum of a finite Geometric series, the sum of row $i$ is
$$
r^i+r^{i+1}+\cdots+ r^n ={r^i-r^{n+1}\over 1-r }.
$$
The sum of the row sums is
$$\eqalign{
\sum_{i=1}^n {r^i-r^{n+1}\over 1-r } &=
{1\over 1-r}\Bigl(\,\sum_{i=1}^n  r^i-   \sum_{i=1}^nr^{n+1}  \Bigr)\cr
&={1\over 1-r}\cdot \biggl({r-r^{n+1}\over 1-r}-nr^{n+1}\biggr)\cr
&={1\over 1-r}\cdot \biggl({r-r^{n+1}\over 1-r}-{(1-r)nr^{n+1}\over1-r}\biggr)\cr 
&={  r-r^{n+1}-(1-r)nr^{n+1}\over (1-r)^2}\cr
&={r-r^{n+1}(1+n-nr)\over (1-r)^2}.
}
$$
A: Hint: Consider $f(x) = \sum_{i=0}^n x^i$ as a function of $x$. This is a geometric series, so you can get a simpler expression for this:
$$\sum_{i=0}^n x^i = f(x) = \frac{1 - x^{n+1}}{1 - x}.$$
Now differentiating $f(x)$ with respect to $x$, we get
$$\sum_{i=1}^n i x^{i-1} = f'(x) = \left(\frac{1 - x^{n+1}}{1 - x}\right)' = \ \color{red}{?}$$
Multiplying everything by $x$ we get
$$\sum_{i=1}^n i x^i = x f'(x) = \ \color{red}{?}$$
Now plug in $x = \frac{1}{2}$ to get
$$\sum_{i=1}^n i \left(\frac{1}{2}\right)^i = \frac{1}{2} f'\left(\frac{1}{2}\right) = \ \color{red}{?}$$
A: 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.
