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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 ?


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When you sum till $n$, the left hand side of your last displayed equation is the sum of $n$ geometric series that grow shorter and shorter. Use the formula for the sum of a finite geometric series on each of the $n$ series, and then sum the sums. – Dilip Sarwate Apr 8 '12 at 13:29
You could also try differentiating $\sum x^{n + 1}$ term by term. Then you will find an expression for $\sum n x^{n}$. Set $x = \frac{1}{2}$ – Pedro Apr 8 '12 at 13:32
up vote 6 down vote accepted

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 $$S(t)=t+2t^2+3t^3+4t^4+ \cdots +(n-1)t^{n-1}+nt^n.$$ Multiply 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.

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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 $$\eqalign{ \sum_{i=1}^n {r^i-r^{n+1}\over 1-r } &= {1\over 1-r}\Bigl(\,\sum_{i=1}^n r^i- \sum_{i=1}^nr^{n+1} \Bigr)\cr &={1\over 1-r}\cdot \biggl({r-r^{n+1}\over 1-r}-nr^{n+1}\biggr)\cr &={1\over 1-r}\cdot \biggl({r-r^{n+1}\over 1-r}-{(1-r)nr^{n+1}\over1-r}\biggr)\cr &={ r-r^{n+1}-(1-r)nr^{n+1}\over (1-r)^2}\cr &={r-r^{n+1}(1+n-nr)\over (1-r)^2}. } $$

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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

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

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