Help with summation: $\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$ How can one evaluate the below sum? Any help would be greatly appreciated.
$$\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$$
 A: $$\sum_{k=1}^\infty\frac{k(k+2)}{15^k}=\sum_{k=1}^\infty\frac{k(k+1)}{15^k}+\sum_{k=1}^\infty\frac{k}{15^k}$$
start with the first term
depending on the geometric series
$$\frac{1}{1-x}=\sum_{k=0}^{\infty }x^k$$
$$\frac{d}{dx}(\frac{1}{1-x})=\sum_{k=1}^{\infty }kx^{k-1}$$
$$\frac{d^2}{dx^2}(\frac{1}{1-x})=\sum_{k=2}^{\infty }k(k-1)x^{k-2}$$
$$=\sum_{k=1}^{\infty }k(k+1)x^{k-1}=\frac{1}{x}\sum_{k=1}^{\infty }k(k+1)x^{k}$$
$$x\frac{d^2}{dx^2}(\frac{1}{1-x})=\sum_{k=1}^{\infty }k(k+1)x^{k}$$
now, for second term
$$\frac{d}{dx}(\frac{1}{1-x})=\sum_{k=1}^{\infty }kx^{k-1}$$
$$x\frac{d}{dx}(\frac{1}{1-x})=\sum_{k=1}^{\infty }kx^{k}$$
$$\sum_{k=1}^\infty\frac{k(k+2)}{15^k}=x[\frac{d^2}{dx^2}(\frac{1}{1-x})+\frac{d}{dx}(\frac{1}{1-x})]$$
then plug $x=\frac{1}{15}$ to get what you need
A: HINT:
Let $T(r)=\dfrac{a+br+cr^2+dr^3}{15^r}$
and $\dfrac{r(r+2)}{15^r}=T(r)-T(r-1)$
Find $a,b,c,d$ comparing the constants and the coefficients of $r,r^2,r^3$
Then use Telescoping Series 
A: If you put an $x$ in your sum:
$$
f(x) = \sum_{k=1}^\infty k(k+2) x^k
$$
what you are looking for is $f(1/15)$. Now the above is a power series. You can make anti-derivatives of it to reduce to a geometric series, of which the sum is known.
A: Let $S=\sum_{k\geq 1}\frac{k(k+2)}{15^k}$. Then:
$$ 14S = 15S-S = 3+\sum_{k\geq 1}\frac{(k+1)(k+3)-k(k+2)}{15^k}=3+\sum_{k\geq 1}\frac{2k+3}{15^k} $$
and if we set $T=\sum_{k\geq 1}\frac{2k+3}{15^k}$ we have:
$$ 14T = 15T-T = 5+\sum_{k\geq 1}\frac{2}{15^k} = 5+\frac{1}{7}$$
hence $T=\frac{18}{49}$ and $\large\color{red}{S=\frac{165}{686}}.$
A: Another trick is the use coefficients of forward differences, $f(k) = k(k+2)$ 
$\begin{matrix}
k & f & \Delta f & \Delta^2 f \cr
1 & 3 & & \cr
2 & 8 & 5 & \cr
3 & 15 & 7 & 2
\end{matrix}$
$$t= \sum_{k=1}^\infty \frac{1}{15^k} = \frac{1/15}{1 - 1/15} = \frac{1}{14}$$
$$ \sum_{k=1}^\infty \frac{k(k+2)}{15^k}= 3t + 5t^2 + 2t^3 = \frac{165}{686}$$ 
Note: above trick assumed k from 1 to $\infty$ 
Source: book "Fundamentals of numerical analysis" by Stephe Kellison, Chapter 6.4.  
A: We can built the formula, bottom up.
Let $s_n = \sum_{k=1}^\infty k^n r^k $ 
$n-(n-1)=1 \Rightarrow (1-r)s_1 = s_0$
$n^2-(n-1)^2=2n-1 \Rightarrow (1-r)s_2 = 2 s_1 - s_0$
$s_0 = r + r^2 + r^3 + \cdots =  \frac{r}{1-r}$
$s_1 = \frac{s_0}{1-r} = \frac{r}{(1-r)^2}$
$s_2 = \frac{2 s_1 - s_0}{1-r} = \frac{r(r+1)}{(1-r)^3}$ 
For $r=\frac{1}{15}, s_0=\frac{1}{14}, s_1=\frac{15}{196}, s_2=\frac{30}{343}$
$$ \sum_{k=1}^\infty \frac{k(k+2)}{15^k} = s_2 + 2 s_1 = \frac{165}{686} $$
