Cube over power of three sum!

I'll first lay out the "plan", as this was slightly lengthy...

We'll call this summation $S = \sum_0^\infty \dfrac{x^3}{3^x}$ , and we'll define a new summation $T = \sum_0^\infty \dfrac{x^2}{3^x}$ .

Let the general term of $S$ be called $a_x$ and that of $T$ be called $b_x$ . Now, we have the following:

$\displaystyle a_{x+1} = \dfrac 1 3 a_x + b_x + c_x + \dfrac 1 {3^{x+1}}$

$\displaystyle b_{x+1} = \dfrac 1 3 b_x + \dfrac 2 3 c_x + \dfrac 1 {3^{x+1}}$ .

where $c_x = \dfrac{ x } {3^x}$

Now we have three steps.

1. To evaluate $\sum_0^\infty c_x$ .

2. To evaluate $T$ .

3. To evaluate $S$ .

*1. To find $\sum_0^\infty c_x$ : *

We know that $\dfrac 1 {1-y} = \sum_{r=0}^\infty y^r$ for $|y| < 1$ . Differentiating this, we have:

$\displaystyle \dfrac 1 {(1-y)^2} = \sum_{r=0}^\infty r y^{r-1}$

Putting $y = \dfrac 13$ , we have:

$\displaystyle ( \dfrac 9 4 )= 3 \sum_{r=0}^\infty \dfrac{r} {3^{r}}$ Thus, $\sum_0^\infty c_x = \dfrac 3 4$ .

*2. To evaluate $T$ : *

Summing up the second equation from $0$ to $\infty$ , we get: $\displaystyle \sum_0^\infty b_{x+1} = \sum_0^\infty \dfrac 1 3 b_x + \sum_0^\infty \dfrac 2 3 c_x + \sum_0^\infty \dfrac 1 {3^{x+1}}$ .

Thus, we have: $\displaystyle T = \dfrac 1 3 T + \dfrac 2 3 \dfrac 3 4 + \dfrac 1 2$ .

Thus we get $T = \dfrac 3 2$ .

*3. To evaluate $S$ : *

Now we have the value of $T$ . So let's sum up the first equation from $0$ to $\infty$ .

$\displaystyle \sum_0^\infty a_{x+1} = \sum_0^\infty \dfrac 1 3 a_x + \sum_0^\infty b_x + \sum_0^\infty c_x + \sum_0^\infty \dfrac 1 {3^{x+1}}$

This is nothing but:

$\displaystyle S = \dfrac 1 3 S + T + \dfrac 3 4 + \dfrac 1 2$

So we have $S = \dfrac {33} 8$ .

Only slightly lengthy, right?! XD