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Question: \(\lim_{n \rightarrow \infty}\frac{1^{99} + 2^{99} + 3^{99} + .... + n^{99}}{n^{100}}\)=...

lim⁡n→∞199+299+399+....+n99n100\lim_{n \rightarrow \infty}\frac{1^{99} + 2^{99} + 3^{99} + .... + n^{99}}{n^{100}}=

A

99100\frac{99}{100}

B

1100\frac{1}{100}

C

199\frac{1}{99}

D

1101\frac{1}{101}

Answer

1100\frac{1}{100}

Explanation

Solution

lim⁡n→∞199+299+399+....+n99n100=lim⁡n→∞∑r=1n(r99n100)\lim_{n \rightarrow \infty}\frac{1^{99} + 2^{99} + 3^{99} + .... + n^{99}}{n^{100}} = \lim_{n \rightarrow \infty}\sum_{r = 1}^{n}\left( \frac{r^{99}}{n^{100}} \right)

=lim⁡n→∞1n∑r=1n(rn)99=∫01x99dx=[x100100]01=1100.= \lim_{n \rightarrow \infty}\frac{1}{n}{\sum_{r = 1}^{n}{\left( \frac{r}{n} \right)^{99} = \int_{0}^{1}{x^{99}dx =}\left\lbrack \frac{x^{100}}{100} \right\rbrack}}_{0}^{1} = \frac{1}{100}.