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.
 A: We have the geometric series
$$
\sum_{k=0}^\infty x^k=\frac1{1-x}
$$
Differentiate to get
$$
\sum_{k=0}^\infty kx^{k-1}=\frac1{(1-x)^2}
$$
Multiply by $x$, and plug in $x=\frac12$.
A: Assume $|x|<1$ and write $$f(x)\equiv\sum_{k=1}^{\infty}kx^k=x\sum_{k=1}^{\infty}kx ^{k-1} = x\left(\sum_{k=1}^{\infty}x ^k\right)' = x\left(\sum_{k=0}^{\infty}x ^k -1\right)'$$
$$=x\left(\frac{1}{1-x}-1\right)' = x\left(\frac{1}{(1-x)^2}\right)
$$
$$=\frac{x}{(1-x)^2}
$$
In you case, $x=\frac12$, so we have the series $$f(\tfrac12)\equiv\sum_{k=1}^{\infty}k\left(\frac12\right)^k=\frac{\frac12}{(1-\frac12)^2}=\boxed{2}
$$
A: \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*}
A: We can try to solve like a geometric series:
$$
S_n = \frac{1}{2} + 2\frac{1}{2^2} + 3\frac{1}{2^3} + \dots + (n - 1)\frac{1}{2^{n - 1}} + n\frac{1}{2^n} \\
\frac{1}{2}S_n = \frac{1}{2^2} + 2\frac{1}{2^3} + 3\frac{1}{2^4} + \dots + (n - 1)\frac{1}{2^{n}} + n\frac{1}{2^{n+1}} \\
S_n - \frac{1}{2}S_n = \frac{1}{2} + (2 - 1)\frac{1}{2^2} + (3 - 2)\frac{1}{2^3} + \dots + (n - (n - 1))\frac{1}{2^n} - n\frac{1}{2^{n+1}}
$$
This gives:
$$
S_n - \frac{1}{2}S_n = \left(\sum_{i = 1}^{i = n} \frac{1}{2^i}\right) - \frac{n}{2^{n+1}} = \frac{1}{2}\left(\sum_{i = 0}^{i = n - 1} \frac{1}{2^i}\right) - \frac{n}{2^{n+1}}
$$
We know the geometric sum: $\lim_{n\rightarrow\infty}\sum_0^n \frac{1}{2^i} = \frac{1}{1 - \frac{1}{2}} = 2$, which gives:
$$
\lim_{n\rightarrow\infty} S_n = \frac{\frac{1}{2}\cdot 2}{1 - \frac{1}{2}} = 2
$$
