Help with summation: $\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$

Question

How can one evaluate the below sum? Any help would be greatly appreciated.

$$\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$$

Answer 1

$$\sum_{k=1}^\infty\frac{k(k+2)}{15^k}=\sum_{k=1}^\infty\frac{k(k+1)}{15^k}+\sum_{k=1}^\infty\frac{k}{15^k}$$ 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

Answer 2

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

Answer 3

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.

Answer 4

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 $\large\color{red}{S=\frac{165}{686}}.$

Answer 5

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.

Answer 6

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 $r=\frac{1}{15}, s_0=\frac{1}{14}, s_1=\frac{15}{196}, s_2=\frac{30}{343}$

$$ \sum_{k=1}^\infty \frac{k(k+2)}{15^k} = s_2 + 2 s_1 = \frac{165}{686} $$