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I'm trying to find the sum of :

$$sum_{i=1}^nfrac i {2^i}$$

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

$$sum_{i=1}^inftyfrac i {2^i} =2$$

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

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 $tne 1$, we have $t+t^2+t^3+ +cdots +t^n=tfrac{1-t^n}{1-t}$.

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.

Let $r=1/2$.

Write all the terms being added as
$$ left.
matrix{
r& phantom{2}r^2& phantom{2}r^3& phantom{2}r^4&cdots&phantom{n}r^n cr
0 & phantom{2}r^2& phantom{2}r^3& phantom{2}r^4&cdots& phantom{2}r^n cr
0&0& phantom{2}r^3& phantom{2}r^4&cdots& phantom{2}r^ncr
& vdots&&&&cr
0&0&0&0&cdots& phantom{2} r^ncr
} right} n-text{rows}
$$
$$
overline{matrix{ r&2r^2&3r^3&4r^4&cdots&nr^n}phantom{dfgfsdfsfs}}
$$
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
$$eqalign{
sum_{i=1}^n {r^i-r^{n+1}over 1-r } &=
{1over 1-r}Bigl(,sum_{i=1}^n r^i- sum_{i=1}^nr^{n+1} Bigr)cr
&={1over 1-r}cdot biggl({r-r^{n+1}over 1-r}-nr^{n+1}biggr)cr
&={1over 1-r}cdot biggl({r-r^{n+1}over 1-r}-{(1-r)nr^{n+1}over1-r}biggr)cr
&={ r-r^{n+1}-(1-r)nr^{n+1}over (1-r)^2}cr
&={r-r^{n+1}(1+n-nr)over (1-r)^2}.
}
$$

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

$$sum_{i=1}^n i left(frac{1}{2}right)^i = frac{1}{2} f'left(frac{1}{2}right) = color{red}{?}$$