Question

If $I=\sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}\frac{k+1}{x(x+1)}dx,$ then

• A. $I>{\log }_{e}99$
• B. $I<{\log }_{e}99$
• C. $I<\frac{49}{50}$
• D. $I>\frac{49}{50}$

Answer 1

Answer: (B), (D)

$I>\frac{49}{50}$

$k=\{1,2,3,...,98\}$
$\text{As}k\le x<k+1$
$\frac{k+1}{x(x+1)}=(k+1)(\frac{1}{x}-\frac{1}{x+1})$
$\therefore \text{Using Sandwich theorem},$
$\frac{1}{x+1}\le \frac{k+1}{x(x+1)}\le \frac{1}{x}$
$\Rightarrow \underset{k}{\overset{k+1}{\int }}\frac{1}{x+1}dx\le \underset{k}{\overset{k+1}{\int }}\frac{k+1}{x(x+1)}dx\le \underset{k}{\overset{k+1}{\int }}\frac{1}{x}dx$

$\Rightarrow \sum _{k=1}^{98}\ln \frac{k+2}{k+1}\le \sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}\frac{k+1}{x(x+1)}\le \sum _{k=1}^{98}\ln \frac{k+1}{k}$
$\Rightarrow \ln (\frac{3}{2}\times \frac{4}{3}\times \frac{5}{4}\times \cdots \times \frac{100}{99})\le \sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}\frac{k+1}{x(x+1)}\le (\frac{2}{1}\times \frac{3}{2}\times \frac{4}{3}\times \cdots \times \frac{99}{98})$
$\Rightarrow \ln 50<I<\ln 99$
$\therefore \frac{49}{50}<I<\ln 99$