Show that if $|z| < 1$ then the series $\sum_{n=0}^{\infty}(n+1)z^{n}$ converges, and find its sum. Show that if $|z| < 1$ then the series $\sum_{n=0}^{\infty}(n+1)z^{n}$ converges, and find its sum.
MY ATTEMPT
The given series indeed converges. This is a consequence of the ratio test:
\begin{align*}
\lim_{n\to\infty}\left|\frac{(n+2)z^{n+1}}{(n+1)z^{n}}\right| = \lim_{n\to\infty}\left(1 + \frac{1}{n+1}\right)|z| = |z| < 1
\end{align*}
Based on it, we may proceed. To begin with, notice that
\begin{align*}
f(z) = \sum_{n=0}^{\infty}(n+1)z^{n} = 1 + 2z + 3z^{2} + 4z^{3} + \ldots
\end{align*}
Due to the properties of power series, we can integrate both sides and switch the integral with the summation operation:
\begin{align*}
F(x) = \int_{0}^{x}f(z)\mathrm{d}z & = \int_{0}^{x}\left(\sum_{n=0}^{\infty}(n+1)z^{n}\right)\mathrm{d}z\\\\
& = \sum_{n=0}^{\infty}\int_{0}^{x}(n+1)z^{n}\mathrm{d}z = x + x^{2} + x^{3} + \ldots = \frac{x}{1-x}
\end{align*}
whenever $|x| < 1$. Since $f$ is continuous in $(-1,1)$, the fundamental theorem of calculus implies that
\begin{align*}
f(x) = F'(x) = \frac{1}{(1-x)^{2}}
\end{align*}
whenever $|x| < 1$, and we are done.
Based on the wording of my solution, I'd like to know if my approach is correct.
Secondly, I'd like to know if there is another way to solve it.
Any contribution is appreciated.
 A: Your approach looks fine; you used a theorem about term-by-term integration along with the FTC, while another way is to use the theorem regarding term-by-term differentiation:

If $F(z)= \sum_{n=0}^{\infty}a_n z^n$ is a power series with radius of convergence $\rho>0$, then:

*

*$F$ is differentiable at every $z$ with $|z| < \rho$

*the "derived series" $f(z) = \sum_{n=0}^{\infty}n a_n z^{n-1}$ also has radius of convergence $\rho$, and $F' = f$. Or more memorably,
\begin{align}
\dfrac{d}{dz}\sum_{n=0}^{\infty}a_n z^n &= \sum_{n=0}^{\infty}\dfrac{d}{dz}(a_nz^n) = \sum_{n=0}^{\infty} na_n z^{n-1} \quad (|z| < \rho)
\end{align}

In your case, we have:
\begin{align}
f(z) &:= \sum_{n=0}^{\infty} (n+1)z^{n} = \sum_{n=0}^{\infty}\dfrac{d}{dz}(z^{n+1})
\end{align}
Now, consider the series $\sum_{n=0}^{\infty}z^{n+1}$; this is a series with radius of convergence $1$, so by the theorem, the derived series also has the same radius of convergence $1$, and
\begin{align}
f(z) &= \dfrac{d}{dz}\sum_{n=0}^{\infty}z^{n+1} = \dfrac{d}{dz}\left(\dfrac{z}{1-z}\right) = \dfrac{1}{(1-z)^2},
\end{align}
where the middle equality is by the geometric series formula.
