Question

${lim}_{\pi \to \infty }\frac{1^{99}+2^{99}+3^{99}+\cdots \cdots n^{99}}{n^{100}}=$ [EAMCET 1994]

• A. $\frac{9}{100}$
• B. $\frac{1}{100}$
• C. $\frac{1}{99}$
• D. $\frac{1}{101}$

Answer 1

The correct option is B

$\frac{1}{100}$

${lim}_{\pi \to \infty }\frac{1^{99}+2^{99}+3^{99}+\cdots \cdots n^{99}}{n^{100}}={lim}_{\pi \to \infty }{\sum }_{r=1}^n\left(\frac{r^{99}}{n^{100}}\right)$
$={lim}_{\pi \to \infty }\frac{1}{n}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^{99}={\int }_0^1{x}^{99}dx={\left[\frac{x^{100}}{100}\right]}_0^1=\frac{1}{100}$