MCQ
$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+{{3}^{99}}+......{{n}^{99}}}{{{n}^{100}}}=$
  • A
    $\frac{9}{{100}}$
  • $ 1/100$
  • C
    $\frac{1}{{99}}$
  • D
    $\frac{1}{{101}}$

Answer

Correct option: B.
$ 1/100$
b
(b) $\mathop {\lim }\limits_{n \to \infty } \,\frac{{{1^{99}} + {2^{99}} + ..... + {n^{99}}}}{{{n^{100}}}} $

$=  \mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{r = 1}^n {\,\left( {\frac{{{r^{99}}}}{{{n^{100}}}}} \right)} $

$ = \mathop {\lim }\limits_{n \to \infty } \,\frac{1}{n}\,\,\sum\limits_{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}}.} } $

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