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\begin{document}
\title{Calculating the Series $1+4r+9r^{2}+\cdots$}
\author{Amit Yaron}
\date{June 14, 2021}

\maketitle
Another one from YouTube:

$1+4r+9r^{2}+\cdots$

The series can be written as:

\[
S(r)=\sum_{n=0}^{\infty}(n+1)^{2}r^{n}
\]

How to solve?

We can derive the series 

\[
\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}
\]

and get

\[
\frac{1}{(1-x)^{2}}=\sum_{n=1}^{\infty}nx^{n-1}=\sum_{n=0}^{\infty}(n+1)x^{n}
\]

derive once again and get:

\[
\frac{2}{(1-x)^{3}}=\sum_{n=1}^{\infty}n(n+1)x^{n-1}=\sum_{n=0}^{\infty}(n+1)(n+2)x^{n}
\]

From the above, we can conclude that we can calculate $S(r)$ without
derivatives and/or anti-derivatives by multiplying it and dividing
it by $(1-r)$ as many times as needed. Let us start:

\begin{align*}
S(r) & =\frac{1-r}{1-r}S(r)=\frac{1}{1-r}\sum_{n=0}^{\infty}(n+1)^{2}(1-r)r^{n}=\frac{1}{1-r}\sum_{n=0}^{\infty}[(n+1)^{2}r^{n}-(n+1)^{2}r^{n+1}]=\\
= & \frac{1}{1-r}[\sum_{n=0}^{\infty}(n+1)^{2}r^{n}-\sum_{n=1}^{\infty}n^{2}r^{n}]=\frac{1}{1-r}\sum_{n=0}^{\infty}[(n+1)^{2}-n^{2}]r^{n}=\frac{1}{1-r}\sum_{n=0}^{\infty}(2n+1)r^{n}=\\
= & \frac{1}{(1-r)^{2}}\sum_{n=0}^{\infty}(2n+1)(1-r)r^{n}=\frac{1}{(1-r)^{2}}\sum_{n=0}^{\infty}[(2n+1)r^{n}-(2n+1)r^{n+1}]=\\
= & \frac{1}{(1-r)^{2}}[\sum_{n=0}^{\infty}(2n+1)r^{n}-\sum_{n=1}^{\infty}(2n-1)r^{n}]=\frac{1}{(1-r)^{2}}[\sum_{n=0}^{\infty}2r^{n}+(0-1)]=\\
= & \frac{1}{(1-r)^{2}}[-1+\sum_{n=0}^{\infty}2r^{n}]=\frac{1}{(1-r)^{2}}(-1+\frac{2}{1-r})=\frac{1+r}{(1-r)^{3}}
\end{align*}

It converges for $r\in(-1,1)$
\end{document}