Tricky limitting sum Find the sum of the following infinite series, given $|x|<1$
$$2+4x+\frac{9}{2}x^2+\frac{16}{3}x^3+\frac{25}{4}x^4+\frac{36}{5}x^5+\frac{49}{6}x^6+\frac{64}{7}x^7+\frac{81}{8}x^8+ \ldots $$
I have tried turning this into a geometric series, but I just didn't even know where to begin. I also tried relating this to well-known sums but that didn't work out either.
I would love some help with this.
 A: The series appears to take the form
$$ f(x) = 2 + \sum_{n=1}^{\infty} \frac{(n+1)^2}{n} \, x^n $$
which can be seen in the form:
\begin{align}
f(x) &= 2 + \sum_{n=1}^{\infty} \frac{(n+1)(n+2)}{n} \, x^n - \sum_{n=1}^{\infty} \frac{(n+1) \, x^n}{n} \\
&= 2 + \frac{d^2}{d x^2} \, \sum_{n=1}^{\infty} \frac{x^{n+2}}{n} - \frac{d}{dx} \, \sum_{n=1}^{\infty} \frac{x^{n+1}}{n} \\
&= 2 + \frac{d^2}{d x^2} \, (- x^2 \, \ln(1-x) ) - \frac{d}{d x} \, (- x \, \ln(1-x) ) \\
&= \cdots \\
&= 2 + \frac{3 x - 2 x^2}{(1-x)^2} - \ln(1-x) \\
&= \frac{2 - x}{(1-x)^2} - \ln(1-x).
\end{align}
A faster way is:
\begin{align}
f(x) &= 2 + \sum_{n=1}^{\infty} \frac{(n+1)^2}{n} \, x^n \\
&= 2 + \sum_{n=1}^{\infty} \frac{n (n+2) + 1}{n} \, x^n \\
&= 2 + \sum_{n=1}^{\infty} \left( (n+2) + \frac{1}{n} \right) \, x^n \\
&= \sum_{n=0}^{\infty} (n+2) \, x^n + \sum_{n=1}^{\infty} \frac{x^n}{n} \\
&= \frac{2-x}{(1-x)^2} - \ln(1-x).
\end{align}
A: So your sum looks like
$$
s(x) = 2 + \sum_{k=1}^\infty \frac{(k+1)^2}{k} x^k.
$$
One approach is to expand the sum to
$$
\begin{split}
\sum_{k=1}^\infty \frac{(k+1)^2}{k} x^k
 &= \sum_{k=1}^\infty \frac{k^2+2k+1}{k} x^k \\
 &= \sum_{k=1}^\infty kx^k
    + 2 \sum_{k=1}^\infty x^k
    + \sum_{k=1}^\infty \frac{x^k}{k}.
\end{split}
$$
The first sum you can get by differentiating a geometric series:
$$
\begin{split}
g(x)    &= \sum_{k=0}^\infty x^k\\
g'(x)   &= \sum_{k=1}^\infty k x^{k-1}\\
x g'(x) &= \sum_{k=1}^\infty k x^k
\end{split}
$$
The second sum is a geometric series. Differentiate the third sum to get a geometric series:
$$
\begin{split}
f(x)  &= \sum_{k=1}^\infty \frac{x^k}{k}\\
f'(x) &= \sum_{k=1}^\infty x^{k-1}
\end{split}
$$
which you can now find...
Just make sure to be careful about the term where $k=0$ everywhere...
A: Note that the series can be written as,
$$2+\sum_{k\geq 1}\frac{(k+1)^{2}}{k}x^{k}$$
$$2+\sum_{k\geq 1}kx^{k}+\frac{x^{k}}{k}+2x^{k}$$
Provided that $|x|<1$ we have that,
$$\sum_{k\geq 1}\frac{x^{k}}{k}=-\ln(1-x)$$
$$\sum_{k\geq 1}kx^{k}=\frac{x}{(1-x)^{2}}$$
And geometric series is already known.
A: Notice that $\frac{(n+1)^2}{n}=n+2+\frac{1}{n}$ so the sum changes to sum of simple series.
