Question
For a positive integer ' $n$ ', let $S(n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{2^{n} - 1}$. Then
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Solution
Sol. [Key Concept: Properties of the partial sums $S_{n}$.]
$$S(n) = 1 + \underset{2\text{~}\text{terms}\text{~}}{\overset{\left( \frac{1}{2} + \frac{1}{3} \right)}{︸}} + \underset{4\text{~}\text{terms}\text{~}}{\overset{\left( \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \right)}{︸}} + \underset{8\text{~}\text{terms}\text{~}}{\overset{\left( \frac{1}{8} + \frac{1}{9} + + \ldots + \frac{1}{15} \right)}{︸}} +$$
$$\cdots\underset{2^{n - 1}\text{~}\text{terms}\text{~}}{\overset{\left( \frac{1}{2^{n - 1}} + \frac{1}{2^{n - 1} + 1}\cdots\frac{1}{2^{n} - 1} \right)}{︸}}$$
$$\begin{matrix} < 1 & \left( \frac{1}{2} + \frac{1}{2} \right) + \left( \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} \right) + \left( \frac{1}{8} + \frac{1}{8} + \ldots + \frac{1}{8} \right) + \ldots + \left( \frac{1}{2^{n - 1}} + \frac{1}{2^{n - 1}} + \ldots\frac{1}{2^{n - 1}} \right) \\ & \ = 1 + 1 + 1 + 1 + \ldots + 1 \\ & \ = n \end{matrix}$$
$$\begin{matrix} & S(n) > \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + \left( \frac{1}{9} + \frac{1}{10} + \ldots + \frac{1}{16} \right) + \ldots \\ & \ + \left( \frac{1}{2^{n - 1} + 1} + \frac{1}{2^{n - 1} + 2} + \ldots\frac{1}{2^{n}} \right) \\ & \ > \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4} \right) + \left( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) + \left( \frac{1}{16} + \frac{1}{16} + \ldots + \frac{1}{16} \right) + \ldots\left( \frac{1}{2^{n}} + \frac{1}{2^{n}} + \ldots + \frac{1}{2^{n}} \right) \end{matrix}$$
$$\begin{matrix} & \ = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \ldots + \frac{1}{2} \\ & \ = \frac{n}{2} \end{matrix}$$
$$\Rightarrow S(2n) > n$$
✔ (a) $S(200) > \frac{200}{2} = 100$
✘ (b) $S(200) > 100$
✔ (c) $S(100) < 100$
✘ (d) $S(100) < 100$
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