Summation problems involving $k c^{-k}$

Question

I am trying to find a better way of solving problems involving summations of the form: $\sum_{k=0}^\infty k\cdot c^{-k}$ and $\sum_{k=0}^\infty k^2 \cdot c^{-k}$

For example, finding $\sum_{k=0}^\infty k\cdot 2^{-k}$, I used the following approach:

$S = \frac{1}{2} + 2 \cdot \frac{1}{2^2} + 3 \cdot \frac{1}{2^3} + ....$
$S = \frac{1}{2}[1+2 \cdot \frac{1}{2} + 3 \cdot \frac{1}{2^2} + .....]$
$S = \frac{1}{2}[1+\sum_{k=0}^\infty (k+1)\cdot 2^{-k}]$
$\sum_{k=0}^\infty (k+1)\cdot 2^{-k} = \sum_{k=0}^\infty k\cdot 2^{-k} + \sum_{k=0}^\infty 2^{-k} = S + \sum_{k=0}^\infty 2^{-k}$

Therefore, $S = 1 + \sum_{k=0}^\infty 2^{-k} = 1 + \frac{1}{1-\frac{1}{2}} = 3$

I may have made a few calculation mistakes (I would appreciate if you could point that out), but my main concern is a simpler way to approach this kind of problems. Can I use this method to evaluate problems of the form $\sum_{k=0}^\infty k^2 \cdot c^{-k}$ ? I tried, but failed.

I would appreciate if someone could point out the most efficient way to solve this

Answer 1

Hint: Observe \begin{align} \frac{1}{1-x} = \sum^\infty_{n=0}x^n \end{align} when $|x|<1$. In particular, we have that \begin{align} \frac{1}{(1-x)^2}=\frac{d}{dx}\frac{1}{1-x} = \sum^\infty_{n=1} nx^{n-1} \ \ \Rightarrow \ \ \frac{x}{(1-x)^2} = \sum^\infty_{n=1}nx^n. \end{align}

Answer 2

Note that we can start with the sum (where I replaced $c$ with $x$) $$\frac{x}{x-1}=\sum_{k=0}^\infty x^{-k}$$ Differentiating and multiplying by $x$, we get that $$\frac{x}{(x-1)^2 }= \sum_{k=0}^{\infty} kx^{-k}$$ Doing this again $$\frac{(x+1)x}{(x-1)^3 }= \sum_{k=0}^{\infty} k^2x^{-k}$$
If you continue this process your denominator will not change, and your numerator will be an $n$th degree polynomial, where $n$ is the power of $k$ on the RHS

Answer 3

The short version:

$$S_2:=\sum_{k=0}^\infty k^2a^{-k}=\sum_{k=1}^\infty(k-1)^2a^{1-k}=a\left(\sum_{k=0}^\infty(k-1)^2a^{-k}-1\right)=a(S_2-2S_1+S_0-1)$$

and

$$S_2=a\frac{2S_1-S_0+1}{a-1}.$$

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The full version:

We can solve this for any power, using

$$S_d:=\sum_{k=0}^\infty k^da^{-k}=\sum_{k=1}^\infty(k-1)^da^{1-k}=a\left(\sum_{k=0}^\infty(k-1)^da^{-k}-(-1)^d\right).$$

Then by the Binomial theorem,

$$(k-1)^d=\binom d0k^{d}-\binom d1k^{d-1}+\binom d2k^{d-2}\cdots+\binom dd(-1)^d,$$

and by subtraction we can cancel the powers $k^d$, giving

$$\left(1-\frac1a\right)S_d=\binom d1S_{d-1}-\binom d2S_{d-2}\cdots+\binom dd(-1)^{d-1}S_0+(-1)^d.$$

In particular

$$\left(1-\frac1a\right)S_0=1,\\ \left(1-\frac1a\right)S_1=S_0-1,\\ \left(1-\frac1a\right)S_2=2S_1-S_0+1,\\ \left(1-\frac1a\right)S_3=3S_2-3S_1+S_0-1,\\ \left(1-\frac1a\right)S_4=4S_3-6S_2+4S_1-S_0+1,\\ \cdots$$

With $a=2$, $$S_0=2,S_1=2,S_2=6,S_3=26,S_4=150,\cdots$$