limn→∞199+299+399+ +n99n100=limn→∞1n199+299+399+ +n99n99=limn→∞1n199n99+299n99+399n99+ +n99n99=limn→∞1n1n99+2n99+3n99+ +nn99=limn→∞1n∑r=1nrn99Using =limn→∞1n∑r=abfrn=∫limn→∞anlimn→∞bnfxdx=∫limn→∞1nlimn→∞nnx99 dx=∫01x99 dx=x10010001=1100

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Solve this: 99 100

\lim_{n \to \infty} \frac{1^{99} + 2^{99} + 3^{99} + . . . . + n^{99}}{n^{100}} = \lim_{n \to \infty} \{\frac{1}{n} (\frac{1^{99} + 2^{99} + 3^{99} + . . . . + n^{99}}{n^{99}})\} = \lim_{n \to \infty} \{\frac{1}{n} (\frac{1^{99}}{n^{99}} + \frac{2^{99}}{n^{99}} + \frac{3^{99}}{n^{99}} + . . . + \frac{n^{99}}{n^{99}})\} = \lim_{n \to \infty} \{\frac{1}{n} ({(\frac{1}{n})}^{99} + {(\frac{2}{n})}^{99} + {(\frac{3}{n})}^{99} + . . . + {(\frac{n}{n})}^{99})\} = \lim_{n \to \infty} \{\frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{99}\} [U \sin g = \lim_{n \to \infty} \{\frac{1}{n} \sum_{r = a}^{b} f (\frac{r}{n})\} = \int_{\lim_{n \to \infty} \frac{a}{n}}^{\lim_{n \to \infty} \frac{b}{n}} f (x) d x] = \int_{\lim_{n \to \infty} \frac{1}{n}}^{\lim_{n \to \infty} \frac{n}{n}} x^{99} d x = \int_{0}^{1} x^{99} d x = {[\frac{x^{100}}{100}]}_{0}^{1} = \frac{1}{100}

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Solve this: Solve this: 99 100

Solve this:

Solve this: 99 100