Summing $\sum_ {k=0}^{\infty} \frac{k^3}{3^k}$ How do I find $\sum_ {k=0}^{\infty} \frac{k^3}{3^k}$ .
I tried like derivative,like I did in other examples,but in this example that doesn't work... Can somebody help? 
 A: $$k^3=k(k-1)(k-2)+3k(k-1)+k.$$
Then by derivation,
$$s\left(\frac13\right):=\sum_{k=0}^\infty\frac1{3^k}=\frac1{1-\frac13},$$
$$s'\left(\frac13\right):=\sum_{k=0}^\infty\frac k{3^{k-1}}=\frac1{\left(1-\frac13\right)^2},$$
$$s''\left(\frac13\right):=\sum_{k=0}^\infty\frac{k(k-1)}{3^{k-2}}=\frac2{\left(1-\frac13\right)^3},$$
$$s'''\left(\frac13\right):=\sum_{k=0}^\infty\frac{k(k-1)(k-2)}{3^{k-3}}=\frac{3!}{\left(1-\frac13\right)^4},$$
and the sum is
$$\frac{3!}{3^3}\left(\frac32\right)^4+\frac{2\cdot3}{3^2}\left(\frac32\right)^3+\frac13\left(\frac32\right)^2= \frac{33}8.$$
A: Suppose you have $$s(x)=\sum x^k$$
Then $$t(x)=xs'(x)=\sum kx^k$$
apply a couple more times, with attention to limits.

So take the lower limit of the sum to be $k=0$ so that $s(x)=\frac 1{1-x}$ with $|x|\lt 1$.
Then $t(x)=xs'(x)$ gives $\sum kx^k$ with lower limit $k=1$, but since the term with $k=0$ is zero, we can take the lower limit as zero. This gives $t(x)=\frac x{(1-x)^2}$.
And $u(x)=xt'(x)$ sums $k^2x^k$ in the same way, with the same comment on limits with $u(x)=\frac {1+x}{(1-x)^3}$.
Now do the same for $v(x)=xu'(x)$.
A: Without differentiation of formal series:
$$S_0:=\sum_{k=0}^\infty\frac1{3^k}=\frac1{1-\frac13}.$$
By shifting of the index and noting that the sums can start at $0$ or $1$ indifferently:
$$S_1:=\sum_{k=0}^\infty\frac k{3^k}=\sum_{k=0}^\infty\frac{k+1}{3^{k+1}}=\frac{S_1+S_0}3,$$
$$S_2:=\sum_{k=0}^\infty\frac{k^2}{3^k}=\sum_{k=0}^\infty\frac{k^2+2k+1}{3^{k+1}}=\frac{S_2+2S_1+S_0}3,$$
$$S_3:=\sum_{k=0}^\infty\frac{k^3}{3^k}=\sum_{k=0}^\infty\frac{k^3+3k^2+3k+1}{3^{k+1}}=\frac{S_3+3S_2+3S_1+S_0}3.$$
This yields
$$S_0=\frac32,
S_1=\frac34,
S_2=\frac32,
S_3=\frac83.$$

Generalizing, we have the simple recurrence
$$S_n(x):=\sum_{k=0}^\infty k^nx^k=\sum_{k=0}^\infty (k+1)^nx^{k+1}=x\sum_{j=0}^n\binom njS_j(x),$$ or
$$S_n(x)=\frac x{1-x}\sum_{j=0}^{n-1}\binom njS_j(x).$$
A: I will write here a curiosity I found in doing this exercise because I find it really amusing. 
Let's test if the series converges:
$$\lim_{\Lambda\to +\infty} \int_0^{\Lambda}\ k^3\cdot 3^{-k}\ \text{d}k = \lim_{\Lambda\to +\infty} \left(- \frac{3^{-k}\cdot (6 + 6k\ln(3) +3k^2 \ln^2(3) + k^3\ln^3(3) )}{\ln^4(3)}\right)_1^{\Lambda}$$
substituting and calculating the limit we obtain
$$\lim_{\Lambda\to +\infty} \int_0^{\Lambda}\ k^3\cdot 3^{-k}\ \text{d}k = \frac{6}{\ln^4(3)} \approx 4.118 $$
The amusing fact is that that series does converge and its sum is
$$\sum_{k = 0}^{+\infty}\ \frac{k^3}{3^k} = \frac{33}{8} = 4.125$$
strictly similar to the integral above.
I'm not such an expert so I demand: is this a funny coincidence or is there some connection (in general)?
P.s. I obtained the sum value with Mathematica, but it's easily computing by hands too.
