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. $a\left(100\right)\le 100$
• B. $a\left(100\right)>100$
• C. $a\left(200\right)\le 100$
• D. $a\left(200\right)>100$

Answer 1

Answer: (A), (D)

$a\left(100\right)\le 100$
$a\left(200\right)>100$

We have

$a\left(n\right)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{2^n-1}$

$=1+(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7})+(\frac{1}{8}+....+\frac{1}{15})+...+\frac{1}{2^n-1}$

$=1+(\frac{1}{2}+\frac{1}{2^2-1})+(\frac{1}{2^2}+\frac{1}{5}+\frac{1}{6}+\frac{1}{2^3-1})+(\frac{1}{2^3}+...+\frac{1}{2^4-1})+...$

$<1+1+...+1=n$

Thus,

$a\left(100\right)<100$

Also,

$a\left(n\right)=1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+...+\frac{1}{2^n-1}$

$=1+\frac{1}{2}+(\frac{1}{2^1+1}+\frac{1}{2^2})+(\frac{1}{2^2+1}+\frac{1}{2^3})+...+(\frac{1}{2^{n-1}+1})$

$>1+\frac{1}{2}+\frac{2}{4}+\frac{4}{8}+....+\frac{2^{n-1}}{2^n}-\frac{1}{2^n}$

$>1+\frac{1}{2}+\frac{2}{4}+\frac{4}{8}+...+\frac{2^{n-1}}{2^n}-\frac{1}{2^n}$

$=1+(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...+\frac{1}{2})-\frac{1}{2^n}$

$=1+\frac{n}{2}-\frac{1}{2^n}=(1-\frac{1}{2^n})+\frac{n}{2}$

Thus,

$a>(1-\frac{1}{2^{200}})+\frac{200}{2}>100$

i.e.,

$a\left(200\right)>100$.