What does the geometric series $ \sum\limits_{n=0}^{\infty} n \cdot \left( \frac{1}{4} \right) ^{n}$ converge to? I am trying to find the value to which the geometric series
$$
\sum\limits_{n=0}^{\infty} n \cdot \left( \frac{1}{4} \right) ^{n},
$$
converges. Now I know how to find a sum when we have a common ratio, but since $n$ is increased with each term, I am stuck at how to tackle this.
Any suggestions would be appreciated.
 A: Beginning with the formula for the sum of a geometric series
$$ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n, \hspace{1cm} |x|<1 $$
and differentiating term by term, we have
$$ \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} nx^{n-1}, \hspace{1cm} |x|<1 $$
justified by absolute convergence. Now set $x = \frac{1}{4} $ and multiply through by $\frac{1}{4}$ to obtain your answer
A: Hint
$$S_n=1\cdot\left( \frac{1}{4} \right) ^{1}+2\cdot\left( \frac{1}{4} \right) ^{2}+...+n\cdot\left( \frac{1}{4} \right) ^{n}$$
Now, multiply by $1/4$
$$\frac{1}{4}S_n=1\cdot\left( \frac{1}{4} \right) ^{2}+2\cdot\left( \frac{1}{4} \right) ^{3}+...+n\cdot\left( \frac{1}{4} \right) ^{n+1}$$
Now subtract,
$$S_n-\frac{1}{4}S_n=1\cdot\left( \frac{1}{4} \right) ^{2}+(2-1)\cdot\left( \frac{1}{4} \right) ^{2}+...+(n-(n-1))\left( \frac{1}{4} \right) ^{n}+n\cdot\left( \frac{1}{4} \right) ^{n+1}$$
Did you understand the idea? Can you finish?
A: A good trick here is to take out one $\frac{1}{4}$.
Lets write
$$\sum_{n=0}^{\infty}n \cdot \left(\frac{1}{4}\right)^n =\sum_{n=1}^{\infty}n \cdot \left(\frac{1}{4}\right)^n = \frac{1}{4}\sum_{n=1}^{\infty}n \cdot \left(\frac{1}{4}\right)^{n-1}$$
Now you can rewrite (but with some caution).
$$\sum_{n=1}^{\infty}n \cdot a^{n-1}=\left(\sum_{n=1}^{\infty} a^{n}\right)'$$
Here if $|a|<1$, you can regard the last term as infinite geometric progression sum. So
$$\left(\sum_{n=1}^{\infty} a^{n}\right)'=\left(\frac{a}{1-a}\right)'=\frac{1-a+a}{(1-a)^2}=\frac{1}{(1-a)^2}$$.
Now just plug in 1/4. And the answer would be
$$\sum_{n=0}^{\infty}n \cdot \left(\frac{1}{4}\right)^n=\frac{1}{4}\cdot\frac{1}{(\frac{3}{4})^2}=\frac{4}{9}$$
If my calculations are correct.
A: The commonly used way is to use the derivative as mentioned in the other answers, which is very helpful for similar, more complicated series. 
However, there is another way you can look at this sum (and this way of looking helps in many calculations in sequences and probability).
You can break up the given sum into the following geometric series: (using the formula you already know) $$\begin{aligned}\dfrac14+\dfrac1{4^2}+\dfrac1{4^3}+\dfrac1{4^4} + \dots &= \dfrac{\frac14}{1-\frac14}=\dfrac43 \cdot\dfrac14\\ \dfrac1{4^2}+\dfrac1{4^3}+\dfrac1{4^4} + \dots &=\dfrac{\frac1{4^2}}{1-\frac14}=\dfrac43\cdot\dfrac1{4^2} \\ \dfrac1{4^3}+\dfrac1{4^4} + \dots &=\dfrac{\frac1{4^3}}{1-\frac14}=\dfrac43\cdot\dfrac1{4^3}\\
\dfrac1{4^4} + \dots &=\dfrac{\frac1{4^4}}{1-\frac14}=\dfrac43\cdot\dfrac1{4^4} \\ \vdots \qquad  & \quad \vdots  \end{aligned} 
$$ since all these terms are positive, adding them in different order wouldn't matter and you might notice that if you add up the sums on the left side of each line, you get ultimately $n$ many $\left(\dfrac14\right)^n$ terms, i.e. one of $\dfrac14$, $2$ of $\dfrac1{4^2}$, $3$ of $\dfrac1{4^3}$ and so on. So the sum of all the terms on the left hand side is the sum you require. 
Now, you see that the terms on the right hand side of each line form a geometric progression with first term $\dfrac{4}{3}\cdot\dfrac14$ and common ratio $\dfrac14$, so that the sum of the terms on the right hand side is $$\dfrac{\dfrac43\cdot \dfrac14}{1-\dfrac14}=\dfrac49$$
