Convergence of the series $\sum\limits_{n=1}^{\infty}nkr^{n-1}$, $|r| < 1$ One series that comes up in actuarial science courses is 
$$\sum\limits_{n=1}^{\infty}nkr^{n-1},\quad |r| < 1\text{, }k>0\text{.}$$
[Typically $|r| < 1$, but I'm wondering if this is a necessary condition.]
This series is a mixed geometric-arithmetic series (I don't know if there's a technical term for it) - how would one go about finding the value of this sum and is $|r|<1$ necessary for convergence?
 A: Consider the series $\sum\limits_{n=0}^{\infty}a_n$ and let 
$$R = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.$$ 
The ratio test states that the series


*

*converges absolutely for $R < 1$,

*diverges for $R > 1$, and

*may or may not converge for $R = 1$.


Applying the ratio test to $a_n = nkr^{n-1}$, you find that $R = |r|$, so the series converges for $|r| < 1$, but you need to check the whether the series converges for those values of $r$ with $|r| = 1$ (i.e. $r = \pm 1$). Checking these two cases, we see that the series diverges (the limit of the terms of a convergent series is zero).
In terms of evaluation, note that $nkr^{n-1} = \frac{d}{dr}(kr^n)$, so $\int nkr^{n-1}dr = kr^n + c$. In light of this, one might hope that we could relate our series to the series with terms $kr^n$ (i.e. a geometric series). It turns out we can do this, because our series is absolutely convergent for $|r| < 1$. In particular, we have
$$\int\left(\sum_{n=1}^{\infty}nkr^{n-1}\right)dr = \left(\sum_{n=1}^{\infty}kr^n\right) + c = \frac{k}{1-r} + c'.$$
The first equality says that we can interchange the integral and the infinite sum; this is where we use absolute convergence. 
So you have an expression for an antiderivative of the series you are interested in. To return back to the original series, we can differentiate both sides with respect to $r$ which gives
$$\sum_{n=1}^{\infty}nkr^{n-1} = \frac{k}{(1-r)^2}.$$
A: Michael's solution is good. I managed to figure this out using a different method with David Mitra's comment. 
Let 
$$S_n = 1 + 2r + 3r^2 + \cdots + nr^{n-1}\text{.}$$
Then
$$rS_n = r + 2r^2 + 3r^3 + \cdots + nr^n\text{,}$$
so that 
$$S_n - rS_n = (1-r)S_n = 1 + r + r^2 + r^3 + \cdots + r^{n-1} - nr^n = \dfrac{1-r^n}{1-r}-nr^n\text{.}$$
This implies that
$$S_n = \dfrac{1-r^n - nr^n + nr^{n+1}}{(1-r)^2}$$
after doing some algebra. By the ratio test, this value converges for $|r| < 1$ (as Michael showed), so that $S_n \to \dfrac{1}{(1-r)^2}$ as $n \to \infty$. Multiplying by $k$ gives the desired sum. 
A: Another way to compute the sum (after check that $R=|r|$ hence the series converges absolutely for $|r|<1$) is the following:
\begin{align*}
\sum_{n=1}^{+\infty}nkr^{n-1}
&=k\sum_{n=1}^{+\infty}nr^{n-1}\\
&=k\sum_{n=1}^{+\infty}(r^{n})'\\
&=k\left(\sum_{n=1}^{+\infty}r^n\right)'\\
&=k\left(\frac1{1-r}-1\right)'\\
&=\frac{k}{(1-r)^2}\;.
\end{align*}
