# How to evaluate $\lim_{n \to \infty} \sum_{k=1}^n \frac{n+k}{n^3+k}$?

How to evaluate the following limit? $$\lim_{n \to \infty} \sum_{k=1}^n \frac{n+k}{n^3+k}$$

I think that every term of the sum is greater than the first one and smaller than the last one and then from the squeeze theorem the limit is $$0$$. However, I can't prove that inequality.

• What are you asking? – Will M. Nov 19 '18 at 18:31
• The sum is less than $n \frac{n +n}{n^3+1}$ – RRL Nov 19 '18 at 18:33

This should work: $$\sum_{k=1}^n\frac{n+k}{n^3+k}\leq\sum_{k=1}^n\frac{n+k}{n^3}\leq\sum_{k=1}^n\frac{n+n}{n^3}=\frac{2n^2}{n^3}=\frac{2}{n}\to0$$
Also, every term is $$\geq 0$$ so $$0$$ is also a lower bound.
• Typo: $n + n = 2n$. – Paul Frost Dec 2 '18 at 10:25
Note that $$n+1\le n+k\le 2n$$ and $$n^3+1\le n^3 +k\le n^3+n$$.