Question

For a positive integer $n,$let $a\left(n\right)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n}\cdot$ Then

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

Answer 1

Answer: (A), (C)

The correct options are
A $a\left(100\right)<100$
C $a\left(200\right)<100$

$a\left(n\right)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{n}$ , where $n\in Z^{+}$
Let us consider a series $b\left(n\right)$
$b\left(n\right)=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots +nterms$
Using comparison test, $b\left(n\right)>a\left(n\right)$ always
$\Rightarrow$ Putting $n=200,b\left(200\right)>a\left(200\right)$
$\Rightarrow 1+\frac{199}{2}>a\left(200\right)\Rightarrow a\left(200\right)<100.5\simeq 100$
Now, let $c\left(n\right)=1+1+1+\cdots +nterms$
$\Rightarrow c\left(n\right)>a\left(n\right)$
Putting $n=100,c\left(100\right)>a\left(100\right)$
$\Rightarrow a\left(100\right)<100$
Hence, all options are incorrect