Question

For a positive integer n let $a (n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ...+ \frac{1}{(2^n)-1},$ then

• A. $a\, (100) \le 100$
• B. $a\, (200) \le 100$
• C. $a\, (100) > 100$
• D. $a\, (200) > 100$

Answer 1

Answer: (D)

$a\, (200) > 100$

Answer 2

Given, $a\, (n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ...+ \frac{1}{(2^n)-1}$
$\, \, \, \, \, = 1 + \bigg(\frac{1}{2} + \frac{1}{3}\bigg) + \bigg(\frac{1}{4} + ...+ \frac{1}{7}\bigg)+\bigg(\frac{1}{8} + ...+ \frac{1}{15}\bigg)$
\hspace50mm + ...+\bigg(\frac{1}{2^{n-1}} + ...+ \frac{1}{2^n - 1}\bigg)
$\, \, \, \, \, < 1 + \bigg(\frac{1}{2} + \frac{1}{2}\bigg) + \bigg(\frac{1}{4}+ \frac{1}{4}+ ...+ \frac{1}{4}\bigg)+\bigg(\frac{1}{8} +\frac{1}{8}+ ...+ \frac{1}{8}\bigg)$
\hspace50mm \, + ...+\bigg(\frac{1}{2^{n-1}} + ...+ \frac{1}{2^n - 1}\bigg)
\hspace20mm \, = 1 + \frac{2}{2} + \frac{4}{4} + \frac{8}{8} + ...+ \frac{2^n-1}{2^n-1}
\hspace20mm \, \underbrace{ = 1 + 1 + 1 + 1 \cdots + 1 = n }_{( n) \, times}
Thus, $a\, (100) \le 100$
Therefore, (a) is the answer.
Again,$a (n) = 1 + \frac{1}{2} + \bigg(\frac{1}{3}+ \frac{1}{4}\bigg)+\bigg(\frac{1}{5} + ...+ \frac{1}{8}\bigg)$
\hspace50mm + ...+ \bigg(\frac{1}{2^{n-1}+1}+...+ \frac{1}{2^n}\bigg)-\frac{1}{2^n}
\hspace20mm > 1 + \frac{1}{2} + \bigg(\frac{1}{4}+ \frac{1}{4}\bigg)+\bigg(\frac{1}{8} +\frac{1}{8}+ ...+ \frac{1}{8}\bigg)
\hspace50mm + ...+ \bigg(\frac{1}{2^n}+...+ \frac{1}{2^n}\bigg)-\frac{1}{2^n}
\hspace20mm = 1 + \frac{1}{2} + \frac{2}{4} + \frac{4}{8} + ...+ \frac{2^{n-1}}{2^n}- \frac{1}{2^n}
\hspace20mm \underbrace{ = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...+ \frac{1}{2} - \frac{1}{2^n} =\bigg(1- \frac{1}{2^n}\bigg)+\frac{n}{2}}_{( n) \, times}
Therefore, $a\, (200) > \bigg(1-\frac{1}{2^{200}}\bigg) + \frac{200}{2} > 100$
Therefore, (d) is also the answer.