# Given k a positive real number, compute lim(1^k + 2^k + 3^k + … + n^k)/n^(k+1) as n approaches infinity.

This is a problem I had on my Math4318 – Analysis II final last Thursday: $\lim_{n\rightarrow\infty}\frac{1^k+2^k+3^k+\cdots +n^k}{n^{k+1}}$

We were also given this hint: “Think about Riemann sums”, but I had no idea what to do with it. I scribbled down a generalized Riemann sum and thought nothing of it. $\int_{a}^{b}{f(x)dx}\approx \sum_{i}{f(t_i)h_i}$

where $h_i$ is the length of a subset of a partition (the “width” of the rectangles) and $t_i$ is some value in the same subset of the partition.

A final exam period lasts for about 3 hours. After I finished most of the other problems, which I took about an hour, I stared at this one blankly for a good half hour, making scribbles and random calculations until I saw the solution. Maybe it was the lack of sleep, or just one of those moments where you just don’t see it, but I just didn’t get it immediately. Shame. But before we get to the solution, some observations:

1. $\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}\leq \frac{n^n+n^n+\cdots+n^n}{n^{k+1}}=1$, so that the limit should be less than one. Just in case we arrive at what we think is a final solution, this can be a the first check.
2. Trying the expression for specific value of k’s. It’s going to depend on how much background knowledge you have. Notice that if
• k=1, then we have: $\frac{1+2+\cdots+n}{n^{2}} = \frac{(\frac{n(n+1)}{2})}{n^2}\rightarrow \frac{1}{2}$
• k=2, then we have: $\frac{1^2+2^2+\cdots+n^2}{n^{3}} = \frac{(\frac{n(n+1)(2n+1)}{6})}{n^3}\rightarrow \frac{1}{3}$
• k=3, and this one’s a stretch of the mind to remember, I sure didn’t on the exam, but I will include it here anyways: $\frac{1^3+2^3+\cdots+n^3}{n^{4}} = \frac{(\frac{n(n+1)}{2})^2}{n^4}\rightarrow \frac{1}{4}$

If the exam taker didn’t remember those special summation formulas, they aren’t necessary to arrive at the answer. But I at least had the first 2, and it’s not too much of a stretch to guess the answer: 1/(k+1). This would have been greatly reinforced with the knowledge of k=3.

To get the final solution, we return to the hint: Riemann sums. Usually, given a continuous function to estimate the integral of, one uses a constant mesh size. Specifically, if n is the number of partitions we are using, and “a” and “b” are the left and right endpoints of the integral, respectively, then the mesh size is (b-a)/n. What if a = 0, and b = 1? Then we have (b-a)/n = 1/n. Then with a little rewriting: $\frac{1^k+2^k+\cdots+n^k}{n^{k+1}} = \frac{1}{n}\left [ \left(\frac{1}{n}\right)^k + \left(\frac{2}{n}\right)^k + \cdots + \left(\frac{n}{n}\right)^k \right ]$

Then we can see the expression we were given in the beginning was really a Riemann sum for the integral of x^k over the interval [0,1], using a constant mesh size over n subintervals, where the approximation rule was to use the right endpoint of the intervals. (To see this, list the points of the partiiton: {0, 1/n, 2/n, 3/n, … , n/n = 1}).

Then as in the classical explanation of integrals to first year university students, if this is an approximation for an integral, as we let the “rectangles” get smaller and smaller (that is, let n approach infinity, so 1/n approaches 0), the approximation gets better and better, ideally approaching the integral itself. In other words, $\lim_{n\rightarrow\infty}\frac{1^k+2^k+3^k+\cdots +n^k}{n^{k+1}} = \int_{0}^{1}{x^k dx} = \left.\frac{x^{k+1}}{k+1}\right|_0^1 = \frac{1}{k+1}$

which fortunately agrees with our conjecture earlier.