# Limit of summation as n goes to infinity

I am trying to solve the following:

Let $q>1$ and $n \in N$. Evaluate $\lim_{n \rightarrow +\infty} \sum_{k=1} ^n \frac{k^{q-1}}{n^q + k^q}$.

I understand that I need to first get the summation into a closed form, and then take the limit of the sum. However, I am not sure how to deal with the summation. I tried to solve the summation in Mathematica (letting k = 6) first so that I knew where to go with solving the summation, but I ended up with this, which I don't now how to work with:

(1) Am I correct in proceeding by finding the closed form of the summation, then taking the limit?

(2) Can someone suggest a strategy for finding the closed form for this summation? I have never worked with an example where I was working with an unidentified real number like this.

• If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this. – gebruiker Mar 27 '16 at 13:10

Write the sum as the Riemann sum of a function: $$\lim_{n\to\infty}\sum_{k=1}^n\frac{k^{q-1}}{n^q + k^q}=\lim_{n\to\infty} \frac1n\sum_{k=1}^n\frac{(k/n)^{q-1}}{1 + (k/n)^q}=\int_0^1\frac{x^{q-1}}{1+x^q}\,dx.$$
• It is just the definition of the Riemann integral using Riemann sums:$$\int_0^1f(x)\,dx=\lim_{n\to\infty}\frac1n\sum_{k=1}^nf(x_k)$$ where $x_k\in[(k-1)/n,k/n]$. Tipical choice is $x_k=k/n$. – Julián Aguirre Jun 24 '15 at 21:58