Derivative of infinite series stuck

Question

I'm trying to solve recurrences having $n^2$ term. I couldn't solve it. But, after researching, I came up with the following equation. How is it done? Could you please explain step-by-step?

$$\sum _{n=0}^{\infty } n^2 z^n = -\frac{z (z+1)}{(z-1)^3} $$

Edit: Why must we write something like $|z|<1$ for it? What does it mean?

Try differentiating the geometric series $\sum\limits_{n=0}^\infty z^{n+2}$, which has convergence radius $|z|<1$, twice and regrouping. — Тyma Gaidash, Dec 17 '23 at 15:59
Also try the method in arithmetico-geometric sequence, to first multiply a partial LHS by $z$ and then subtract, to reduce to something like $\sum\left[n^2-(n-1)^2\right] z^n = \sum(2n-1)z^n = 2\sum nz^n - \sum z^n$. Then both $\sum nz^n$ and $\sum z^n$ are known. — peterwhy, Dec 17 '23 at 16:03
@ТymaGaidash why should we insist on saying which has convergence radius |z|<1? Why is it important? — Soner from The Ottoman Empire, Dec 17 '23 at 16:03
@SonerfromTheOttomanEmpire Because for $z$ not satisfying $|z|\geq 1$, the left and right hand sides are not equal. — K.defaoite, Dec 17 '23 at 16:19
Ordinarily, an infinite sum $\sum_{n=0}^\infty a_n$ means the limit of finite sums $\sum_{n=0}^N a_n$ as $N$ goes to $\infty$. If $|z|$ is greater than the radius of convergence, this limit does not exist. — Robert Israel, Dec 17 '23 at 16:20
Related to Find a closed form of the series $\sum_{n=0}^{\infty} n^2x^n$. — peterwhy, Dec 17 '23 at 16:25
I suspected this was a duplicate but the search function is not the best... — whpowell96, Dec 17 '23 at 17:05

whpowell96 · Answer 1 · 2023-12-17T16:11:34.397

We being from the well-known geometric series and differentiate $$\begin{aligned} \sum_{n=0}^\infty z^n &= \frac{1}{1-z}\\ \implies \sum_{n=0} nz^{n-1} &= \frac{1}{(1-z)^2}. \end{aligned} $$ We will differentiate again to get $n^2$ in our series, then add a correction to get the final answer. $$ \begin{aligned} \sum_{n=0} nz^{n-1} &= \frac{1}{(1-z)^2}\\ \implies \sum_{n=0} n(n-1)z^{n-2} &= \frac{2}{(1-z)^3}. \end{aligned} $$ To get this to the series we want, we need to correct the powers from $n-2$ to $n$ and correct the coefficients from $n(n-1)$ to $n^2$. The first step is done by simply multiplying through by $z^2$. The second step is done by adding $\sum_{n=0}^\infty nz^n$. However, notice that this is just $z$ times the first derivative we calculated, which is $\frac{z}{(1-z)^2}$. Combining these, we have $$ \sum_{n=0}^n n^2 z^n = \frac{2z^2}{(1-z)^3} + \frac{z}{(1-z)^2} = \frac{z(z+1)}{(1-z)^3}, $$ which is equal to your expression.

As for why we must restrict ourselves to $|z|<1$, this is because the original expression for the gemoetric series (and hence all subsequent calculations) is only valid for these values of $z$. One can easily see that if $|z|\geq 1$, these sums will not converge.

How is The second step is done by adding $\sum{n=0}^\infty nz^n$._ done? — Soner from The Ottoman Empire, Dec 17 '23 at 17:00
We want $n^2 z^n$. We have $(n^2 - n)z^n$ after multiplying through by $z^2$. If we look back through our calculation, we have $nz^{n-1}$, so we multiply that by $z$ and add. — whpowell96, Dec 17 '23 at 17:04

Derivative of infinite series stuck

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