<s>[INST] Simplification of geometric series. Can someone please help simplify this series?
$$\sum_{k=1}^\infty k\left(\frac 12\right)^k$$
In general, $$\sum_{k=1}^\infty\left(\frac 12\right)^k = \frac{1}{1-\frac{1}{2}} =2.$$
However, I am confused with the $k$ in front of the term $k\big(\frac 12\big)^k$.
I understand if the problem is $\sum_{k=1}^\infty\big(\frac 12\big)^k$. However, it is the $k$ term that I don't understand. The answer is suppose to be $2$.
Can someone help me? Thank you.
 [/INST] \begin{align*}
\sum_{k=1}^\infty k\left(\frac{1}{2}\right)^k &= \frac{1}{2} + 2 \frac{1}{2^2} + 3 \frac{1}{2^3} + 4 \frac{1}{2^4} \dots\\
&= \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \right) + \left(\frac{1}{2^2} + 2 \frac{1}{2^3} + 3 \frac{1}{2^4} + \dots\right)\\
&= \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \right) + \left(\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \dots \right) + \left(\frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^4} + \dots \right) + \dots\\
&= 1 + \frac{1}{2} + \frac{1}{2^2} + \dots\\
&= 2
\end{align*}
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