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

Question

Evaluate $\lim\limits_{n\to\infty} \dfrac{1^{99} + 2^{99} + \cdots + n^{99}}{n^{100}}$

This is the question I remembered from my high school textbook (I remembered it while reading about something related). Now I know how to do it using Faulhaber's formula (at least the first coefficient, which is required). But the textbook used this method which I didn't and don't understand. $$\lim\limits_{n\to\infty} \dfrac{1^{99} + 2^{99} + \cdots + n^{99}}{n^{100}} = \lim\limits_{n\to\infty}\dfrac{\large\int\limits_{0}^n x^{99} dx}{n^{100}} = \lim\limits_{n \to \infty}\dfrac{\dfrac{n^{100}}{100}}{n^{100}}= \dfrac{1}{100}$$

What I don't understand how is converting from sum to integral is justified (the textbook didn't justify) and when is it allowed to replace sum by integral?

Answer 1

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

Answer 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.)

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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

Answer 3

See Euler-Maclaurin formula. In this case it looks like $$\sum\limits_{k=0}^nk^{99}=\int\limits_0^nx^{99}dx+\sum\limits_{i=1}^{\infty}\frac{B_i99!}{i!(100-i)!}n^{100-i}$$ Second summand is actually a polynomial of degree $99$, so it is $o(n^{100})$ and $$\lim\frac{\sum\limits_{k=0}^nk^{99}}{n^{100}}=\lim\frac{\int\limits_0^nx^{99}dx}{n^{100}}$$

Answer 4

In general $$ \frac{1^a+2^a+\cdots+n^a}{n^{a+1}}\to\frac{1}{a+1}, $$ for every $a>-1$, as $$ \frac{1^a+2^a+\cdots+n^a}{n^{a+1}}=\frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^a\to\int_0^1x^a\,dx=\frac{1}{a+1}, $$ To make it easier to understand $$ \frac{1}{n}\left(\frac{k}{n}\right)^a-\int_{\frac{k-1}{n}}^{\frac{k}{n}}x^a\,dx=\int_{\frac{k-1}{n}}^{\frac{k}{n}} \left(\left(\frac{k}{n}\right)^a-x^a\right)\,dx. $$ But, using Mean Value Theorem $$ \left(\frac{k}{n}\right)^a-x^a=\left(\frac{k}{n}-x\right)a\xi^{a-1}, $$ for some $\xi\in(\frac{k-1}{n},\frac{k}{n})$, and thus $$ 0\le \left(\frac{k}{n}\right)^a-x^a\le a\frac{1}{n}\left(\frac{k}{n}\right)^{a-1}, $$ and $$ 0\le\int_{\frac{k-1}{n}}^{\frac{k}{n}} \left(\left(\frac{k}{n}\right)^a-x^a\right)\,dx\le a\frac{1}{n^2}\left(\frac{k}{n}\right)^{a-1}\le \frac{a}{n^2}, $$ and finally $$ 0\le \frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^a-\int_0^1x^a\,dx\le \frac{a}{n}\to 0. $$

Answer 5

Replacing a sum with an integral can be justified, but you actually need to justify it. It looks like your textbook didn't. For example, if the function involved something like $\sin kπ$ or $\cos 2kπ$ then for integer values of k you always get a result of 0 or 1, but for real values of k the result can be all over the place, and replacing a sum with an integral would be absolutely not justified.

If you take the constant number $k^{99}$, and compare it with the integral of $x^{99}$ taken from k-1 to k, then you are integrating over values that are all less than $k^{99}$, so the integral is less. If you compare with the integral of $x^{99}$ taken from k to k+1, you are integrating over values that are all greater than $k^{99}$, so the integral is greater.

Now you are calculating the sum of $k^{99}$ for k from 1 to n. We compared each of these n numbers with an integral, so you can also compare the sum with an integral: The whole sum is larger than the integral of $x^{99}$ from 0 to n, and less than the integral from 1 to n+1. That argument would work for any function that is increasing: If a function is increasing for real x, then the sum from 1 to n is between the integral from 0 to n-1 and the integral from 1 to n.

In this case, we can calculate the integrals and find that the sum is between $n^{100}/100$ and $((n+1)^{100} - 1)/100$ which is enough to prove the limit. Your textbook just took the limit for the integral from 0 to n. That's not enough; if you had a function where taking the limit from 1 to n+1 would give a different result, it would be wrong. So that's another thing that can be justified, but must actually be justified.

Answer 6

I would start by observing that $k^m = m!{k\choose m} + P(k)$, where $P$ is a polynomial of degree $<m$.

We will show, by induction, that $\sum_{k=1}^n k^m$ is a polynomial in $n$ with highest degree term $\frac{1}{m+1}n^{m+1}$. This is clear for $m=0$.

We have $\sum_{k=1}^n P(k)$ is a polynomial in $n$ of degree $<m+1$, by the inductive hypothesis. Also $\sum_{k=1}^n m!{k\choose m} = m!{{n+1}\choose{m+1}}$, which has highest degree term $\frac{1}{m+1}n^{m+1}$—and we're done.

To compute the limit, we can now write: $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^{99}} {n^{100}} = \lim_{n\to\infty}\frac{\frac{1}{100}n^{100} + \cdots} {n^{100}} = \frac{1}{100}$$