Show that $\sum_{k=0}^{\infty} \frac{k^2+3k+2}{2} z^k = \frac{1}{(1-z)^3}$, without using differentiation

Question

Show that,

$$\sum_{k=0}^{\infty} \frac{k^2+3k+2}{2} z^k = \frac{1}{(1-z)^3}$$

where $ z \in \mathbb{C}, |z|< 1$

Well, I have figured out that is a Laurent series I have watched 3 videos in the topics but I still don't know how to prove it. I appreciate any kind of help a lot.

Answer 1

FYI,

Based on Newton Binomial theorem known that:

$\frac{1}{(1-z)^{\alpha+1}}=\sum\limits_{k=0}^\infty \binom{k+\alpha}{k}z^k$

In our case $\alpha=2$

$\sum\limits_{k=0}^\infty \frac{(k+2)!}{k!2!}z^k=\sum\limits_{k=0}^\infty \frac{(k+1)(k+2)}{2}z^k$

Answer 2

hint...consider the sum of an infinite geometric series and differentiate it twice

Answer 3

$$\sum_{k=0}^{\infty} \frac{k^2+3k+2}{2} z^k$$ $$\Rightarrow \frac 12\sum_{k=0}^{\infty}k^2z^k+\frac 32\sum_{k=0}^{\infty}kz^k+\sum_{k=0}^{\infty}z^k$$ $$\Rightarrow\frac 12A+\frac 32B+C$$
$$A=z+2^2z^2+3^2z^3+4^2z^4\cdots(a)$$ $$\Rightarrow zA=z^2+2^2z^3+3^2z^4+4^2z^5\cdots(b)$$ subtract $b$ from $a$, $$(1-z)A=z+3z^2+5z^3+7z^4\cdots(c)$$ $$\Rightarrow z(1-z)A=z^2+3z^3+5z^4+7z^5\cdots(d)$$ subtract $d$ from $c$, $$(1-z)^2A=z+2(z^2+z^3+z^3+\cdots)$$ $$\Rightarrow (1-z)^2A=z+2(\frac{z^2}{1-z})$$ $$\Rightarrow A=\frac{z+z^2}{(1-z)^3}$$ Similarly, you can find $B$ and $C$ and finally plug the values in our original expression to get the final result!

Answer 4

You can verify in the infinite series $\sum_k\frac{(k+1)(k+2)}{2}z^k(1-3z+3z^2-z^3)$ the constant term is $1$, and there are no overall $z$ and $z^2$ terms. For $n\ge3$, the $z^n$ coefficient is$$\begin{align}\tfrac{(n+1)(n+2)-3n(n+1)+3n(n-1)-(n-1)(n-2)}{2}&=\tfrac{(n+1)(2-2n)+(n-1)(2n+2)}{2}\\&=-(n+1)(n-1)+(n-1)(n+1)\\&=0.\end{align}$$The $|z|<1$ constraint results from $\sum_k\frac{(k+1)(k+2)}{2}z^k$ having radius of convergence $1$.