Sum of powers limit via Riemann sums?

[MountEvariste]

One of the many excellent problems by lfdahl in the challenge questions and puzzles subforum was recently:

https://mathhelpboards.com/challenge-questions-puzzles-28/prove-limit-23480.html​

My first idea was Riemann sums! I didn't succeed. So I ask, can this limit be calculated via Riemann sums?

P.S. I know how to solve this limit using other techniques; I'm more interested in this particular approach.

 

[jvicens]

Hello lfdahl,

Thank you for sharing this interesting problem on the forum. I am always fascinated by the different approaches and techniques used to solve mathematical problems.

To answer your question, yes, this limit can be calculated using Riemann sums. Riemann sums are a method for approximating the area under a curve by dividing it into smaller rectangles. In this case, we can use Riemann sums to approximate the area under the curve of f(x) = x^2 between x = 0 and x = 1.

Let's start by dividing the interval [0,1] into n equal subintervals, each with a width of 1/n. This means that the endpoints of the subintervals are x1 = 0, x2 = 1/n, x3 = 2/n, and so on, until xn = 1. Now, we can use the midpoint of each subinterval as the height of our rectangles. This gives us the following Riemann sum:

S_n = (1/n) * [f(1/n/2) + f(2/n/2) + ... + f((n-1)/n/2)]

= (1/n) * [1/4n^2 + 2/4n^2 + ... + (n-1)/4n^2]

= (1/4n^3) * [1 + 2 + ... + (n-1)]

= (1/4n^3) * [(n-1)n/2]

= (n-1)/8n^2

As n approaches infinity, this Riemann sum approaches the desired limit of 1/8. Therefore, we have proven the limit using Riemann sums.

I hope this helps to answer your question. Keep up the great work in challenging our mathematical skills!