Evaluate $\sum_{k=1}^\infty k \alpha^k$ I'm struggling with the derivation of this known result. As a prerequisite, consider the following results. Suppose we have time-series data {$x_i$ for $i = 1, \dots, N$} from a stationary process having mean $\mu$ and variance $\sigma^2$ and theoretical acf $\rho(k)$. The sample mean is $\bar{X} = \sum_{i=1}^N \frac{X_i}{N}$. For correlated observations, it can be shown that the sample mean variance is not $\sigma^2/N$ but is given by:
$$
Var(\bar{X}) = \frac{\sigma^2}{N}\left[ 1 + 2 \sum_{k=1}^{N-1}\left( 1 - \frac{k}{N} \right) \rho(k) \right]
$$
I'd like to find the variance of the sample mean for a stationary AR(1) process. The book states that $Var(\bar{X})$ for AR(1) reduces to $ Var(\bar{X}) = \frac{\sigma^2}{N}\left( \frac{1+\alpha}{1-\alpha} \right)$. To show this, I start with the autocorrelation function of an AR(1) process with parameter $\alpha$ which is $\rho(k) = \alpha^{|k|}$ where $\alpha \lt 1$. So the above equation becomes
$$
Var(\bar{X}) = \frac{\sigma^2}{N}\left[ 1 + 2 \sum_{k=1}^{N-1}\left( 1 - \frac{k}{N} \right) \alpha^k \right]
$$
Then let $N \rightarrow \infty $. The complication then is ultimately in solving $\sum_{k=1}^\infty (1 - \frac{k}{N}) \alpha^{|k|}$. If I split it then:
$$
\sum_{k=1}^\infty \alpha^k = \sum_{k=0}^\infty \alpha^k - 1 = \frac{\alpha}{1-\alpha}
$$
Now I need to solve the second part which is where I get stuck. I don't know how to evaluate:
$$
\sum_{k=1}^\infty k \alpha^k
$$
Ultimately, I'd like to show that $ Var(\bar{X}) = \frac{\sigma^2}{N}\left( \frac{1+\alpha}{1-\alpha} \right)$ but I believe by solving the above sum which I'm stuck on I'll be able to show this result from the general equation for $Var(\bar{X})$ above.
 A: Either take the derivative approach as Kenta S mentioned, or do the following:
$$S=\sum_{1}^{\infty} k\alpha^k  \\ \alpha S= \sum_1^{\infty} k\alpha^{k+1} = \sum_2^{\infty} (k-1)\alpha^k =\sum_{2}^{\infty} k\alpha^k -\sum_2^{\infty} \alpha^k \\ =(S-\alpha) -\frac{\alpha^2}{1-\alpha} \\ \implies S= \frac{\alpha}{1-\alpha} +\left ( \frac{\alpha}{1-\alpha} \right)^2$$
A: $$\sum_{k=1}^{\infty} k\alpha^k$$
$$=\sum_{i=1}^{\infty}\sum_{k=i}^{\infty} \alpha^k$$
$$=\sum_{i=1}^{\infty} \alpha^i \sum_{k=0}^{\infty} \alpha^k$$
$$=\sum_{i=1}^{\infty} \alpha^i \frac{1}{1-\alpha}$$
$$=\frac{1}{1-\alpha} \sum_{i=1}^{\infty} \alpha^i $$
$$=\frac{\alpha}{(1-\alpha)^2}$$
A: Comments for the save.
$$\sum_{k=1}^\infty k \alpha^k=\frac{\partial}{\partial \alpha}\alpha^2\sum_{k=0}^\infty  \alpha^{k}-\alpha \sum_{k=0}^\infty\alpha^k$$
which is
$$=\frac{\partial}{\partial \alpha}\frac{\alpha^2}{1-\alpha}-\frac{\alpha}{1-\alpha}$$
$$=(\frac{\alpha}{1-\alpha})^2+\frac{\alpha}{1-\alpha}$$
A: So I answer it this way. This answer assumes $|a| < 1$.
$$
\sum_{k=1}^\infty k a^k = \sum_{k=0}^\infty k a^k = \sum_{k=0}^\infty a \frac{d}{d a} [a^k] = a \sum_{k=0}^\infty \frac{d}{da}[a^k] = a \frac{d}{da}[\sum_{k=0}^\infty a^k] = a \frac{d}{da} [\frac{1}{1-a}] = \cdots
$$
