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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 ?

Regards

  • 1
    See also answers to these questions: [How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$?](http://math.stackexchange.com/questions/11464/how-to-compute-the-formula-sum-limits-r-1d-r-cdot-2r), [What is the limit of $\sum \limits_{n=1}^{\infty}n(2/3)^n$](http://math.stackexchange.com/questions/90637/what-is-the-limit-of-sum-limits-n-1-inftyn2-3n), [How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$](http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-1-infty-frac2n3n1).2012-04-08

3 Answers 3

13

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.

6

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}. } $

4

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}{?}