Question

• Evaluate \(\displaystyle\sum^{\infty}_{n = 1}\frac{n}{2^{n-1}}\).

Solution

$$ \begin{align*} S = \sum^{\infty}_{n = 1}\frac{n}{2^{n-1}} &= \frac 1 1 + \frac 2 2 + \frac 3 4 + \frac 4 8 \cdots \\ &= 1 + \frac 1 2 + \frac 1 2 + \frac 1 4 + \frac 2 4 + \frac 1 8 + \frac 3 8 \cdots \\ &= (1 + \frac 1 2 + \frac 1 4 + \frac 1 8 \cdots) + (\frac 1 2 + \frac 2 4 + \frac 3 8 + \cdots) \\ &= 2 + \sum^{\infty}_{n=1} \frac{n}{2^n} \\ &= 2 + \frac{S}{2} \\ \end{align*} \\ S = 2 + \frac{S}{2} \\ S = 4 $$