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??
Thanks for your help :)
 A: 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 $t^4$
A: 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)
