# Proof that $\sum \limits_{k=3}^\infty k\left({\frac{5}{6}}\right)^{k-3} =48$

I have this problem

Proof that

$\displaystyle\sum_{k=3}^\infty k\left({\displaystyle\frac{5}{6}}\right)^{k-3} =48$

I tryed this:

$\displaystyle\sum_{k=3}^\infty k\left({\displaystyle\frac{5}{6}}\right)^{k-3} = \displaystyle\sum_{k=1}^\infty k\left({\displaystyle\frac{5}{6}}\right)^{k-3} - \displaystyle\sum_{k=1}^2 k\left({\displaystyle\frac{5}{6}}\right)^{k-3} = \displaystyle\sum_{k=1}^\infty k\left({\displaystyle\frac{5}{6}}\right)^{k-3}-3.48$, Now

$\displaystyle\sum_{k=1}^\infty k\left({\displaystyle\frac{5}{6}}\right)^{k-3} = 1*\left({\displaystyle\frac{5}{6}}\right)^{-2} + 2*\left({\displaystyle\frac{5}{6}}\right)^{-1}+ 3*\left({\displaystyle\frac{5}{6}}\right)^{0} + ......... = \left({\displaystyle\frac{5}{6}}\right)^{-2}(1+2*\left({\displaystyle\frac{5}{6}}\right)^{1}+ 3*\left({\displaystyle\frac{5}{6}}\right)^{2} + .....)$

Let $x=(5/6)$

\begin{eqnarray} 1+2x+3x^2+4x^3+\dots = 1 + x + x^2 + x^3 + \dots \\ + x + x^2+ x^3 + \dots\\ + x^2 + x^3 + \dots \\ +x^3 + \dots \\ + \dots \\ =1 + x + x^2 + x^3+\dots \\ +x(1+x+x^2+\dots) \\ +x^2(1+x+\dots)\\ +x^3(1+\dots)\\ +\dots \\ =(1+x+x^2+x^3+\dots)^2=\frac{1}{(1-x)^2} \end{eqnarray}

then

$\left({\displaystyle\frac{5}{6}}\right)^{-2}\left({\displaystyle\frac{1}{1-5/6}}\right)^{2}= 51.48$ finally

$\displaystyle\sum_{k=3}^\infty k\left({\displaystyle\frac{5}{6}}\right)^{k-3} =51.48-3.84 = 48$

Am I right??

HINT:

If we are allowed to use Calculus,

we know $$\sum_{0\le r<\infty}a t^r=\frac a{1-t}$$ if $|t|<1$

Differentiate both sides wrt $t$ and then multiply by $t^{-4}$

Alternatively, we can use Arithmetico-geometric series formula to show that $$\sum_{0\le r<\infty}rat^r=\frac a{1-r}+\frac r{(1-r)^2}$$

Now divide either sides by $t^4$

• On your second last line, you need $S(1-x)$ instead of just $(1-x)$. – Andrew D Sep 28 '13 at 18:02

The crux of your proof -- that

$$\sum_{n=0}^{+\infty} n x^{n-1}$$

factors as

$$\sum_{n=0}^{+\infty} (n+1) x^{n} = \left( \sum_{n=0}^{+\infty} x^{n} \right)^2$$

does indeed look correct.

It might be good to translate your argument into ordinary summation notation:

\begin{align} \sum_{n=0}^{+\infty} (n+1) x^{n} &= \sum_{n=0}^{+\infty} \sum_{m=0}^n x^{n} \\&= \sum_{m=0}^{+\infty} \sum_{n=m}^{+\infty} x^n \\&= \sum_{m=0}^{+\infty} x^m \sum_{n=0}^{+\infty} x^n \\&= \left(\sum_{m=0}^{+\infty} x^m\right) \left( \sum_{n=0}^{+\infty} x^n \right) \end{align}

Can you see how your argument corresponds to this calculation?

(this sort of calculation doesn't work in general: specifically where I swapped the order of the two sums. But power series, geometric series, and sums of positive numbers are all very nice and allow natural things like this to work)