Evaluating $\lim_{n\to\infty} \frac{1^{99} + 2^{99} + \cdots + n^{99}}{n^{100}}$ using integral

Solution 1:

$$\displaystyle \lim_{n\to \infty}\sum^{n}_{k=1}\big(\frac{k}{n}\big)^{99}\cdot\frac{1}{n}=\int^1_0x^{99}dx\,=\frac{1}{100}$$

Solution 2:

Since for $x\in(k,k+1)$ we have $$k^{99} \le x^{99} \le (k+1)^{99}$$ we get $$k^{99} = \int_k^{k+1} k^{99} \,\mathrm{d}x \le \int_k^{k+1} x^{99}\,\mathrm{d}x \le \int_k^{k+1} (k+1)^{99} \,\mathrm{d}x = (k+1)^{99}.$$ This yields $$\sum_{k=0}^{n-1} k^{99} \le \int_0^{n} x^{99}\,\mathrm{d}x \le \sum_{k=0}^{n-1} (k+1)^{99}\\ 1^{99}+2^{99}+\dots+(n-1)^{99} \le \int_0^{n} x^{99}\,\mathrm{d}x \le 1^{99}+2^{99}+\dots+{n}^{99}.$$ Dividing by $n^{100}$ we get $$\frac{1^{99}+2^{99}+\dots+(n-1)^{99}}{n^{100}} \le \frac{\int_0^{n} x^{99}\,\mathrm{d}x}{n^{100}} \le \frac{1^{99}+2^{99}+\dots+{n}^{99}}{n^{100}}.$$

Since $$\left(\frac{1^{99}+2^{99}+\dots+{n}^{99}}{n^{100}}-\frac{1^{99}+2^{99}+\dots+(n-1)^{99}}{{n}^{100}}\right) = \frac1n \to 0$$ we get that all three expression above have the same limit for $n\to\infty$. (Provided that the limit exists at least for one of them.)

What we did above is basically comparing integral (area under the curve) with a sum (area given by the steps in the following image):

I have taken this picture from this answer.

Basically the same derivation is given in this answer