Question

Sol. [Key Concept: Evaluate the limit / sum in closed form.]

$$S = \sum_{n = 1}^{\infty}\mspace{2mu}\sum_{k = 1}^{n - 1}\mspace{2mu}\frac{k}{2^{n + k}} = \sum_{k = 1}^{\infty}\mspace{2mu}\left( \frac{k}{2^{k}}\sum_{n = k + 1}^{\infty}\mspace{2mu}\frac{1}{2^{n}} \right) = \sum_{k = 1}^{\infty}\mspace{2mu}\frac{k}{2^{k}} \times \frac{1}{2^{k + 1}} \times 2$$

$$\sum_{k = 1}^{\infty}\mspace{2mu}\frac{k}{4^{k}} = \frac{4}{9}$$

✘ (a) not equal to the sum

✔ (b) $S = \sum_{k = 1}^{\infty}\frac{k}{4^{k}} = \frac{49}{}$

✔ (c) equals the intermediate form $\sum_{m = 1}^{\infty}\frac{m}{2^{m}}\sum_{n = m + 1}^{\infty}\frac{1}{2^{n}}$

✘ (d) $S = \frac{49}{\neq}\frac{4}{27}$