Calculate $\sum\limits_{n=1}^{\infty}na(1-a)^{n-1}$, where $a \in (0,1)$. Calculate $a\displaystyle \sum_{n=1}^{\infty}n(1-a)^{n-1}$
where $a \in (0,1)$.
 A: Hint: Consider the series:
$$\sum_{n=0}^{\infty} b^n = \frac{1}{1-b}$$
And take the derivative of both sides with respect to $b$. Now apply this to your series.
A: Hints:
$$\forall\,x\in\Bbb C\;,\;|x|<1\;,\;\;f(x):=\frac1{1-x}=\sum_{n=0}^\infty x^n\implies$$
$$f'(x)=\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}\ldots\ldots$$
A: For fun (and this is probably not the solution you are expected to give) we give a mean proof. A certain coin has probability $a$ of landing heads, and probability $1-a$ of landing tails. The coin is tossed repeatedly. Let random variable $X$ be the number of tosses until we get the first head. 
The probability that $X=1$ is $a$, the probability that $X=2$ is $(1-a)a$, the probability that $X=3$ is $(1-a)^2a$, and so on. It follows that the infinite sum we are asked to evaluate is $E(X)$, the mean of $X$. 
Let us calculate this mean another way. Let the mean be $\mu$. Toss the coin.  With probability $a$, we get a head on the first toss, and $X=1$. With probability $1-a$ we get a tail on the first toss, and we have wasted a toss. The mean number of additional tosses needed in that case is $\mu$. Thus
$$\mu=(a)(1)+(1-a)(1+\mu).$$
Solve this linear equation for $\mu$. It turns out that $\mu=\dfrac{1}{a}$.
