Direct evaluation of Limit of a sum

Question

I was trying to solve this problem: $$L=\displaystyle\lim_{n\to\infty }{\sum_{r=1}^{n}\dfrac{r}{n^2+n+r}}$$

I managed to prove that it is $\tfrac{1}{2}$ using the Sandwich theorem. Is there some way, to either:

1. Convert this into an integral
2. Evaluate this sum directly.

Answer 1

$$S_n={\sum_{r=1}^{n}\dfrac{r}{n^2+n+r}}=n(n+1) \left(H_{n(n+1)}-H_{n (n+2)}\right)+n$$

Using the asymptotics of harmonic numbers $$S_n=\frac{1}{2}-\frac{1}{3 n}+\frac{5}{12 n^2}+O\left(\frac{1}{n^3}\right)$$

Answer 2

We have $$n\int_0^1\frac x{n+1+x}\mathrm dx\leq\sum_{r=1}^n\frac r{n^2+n+r}\leq\frac 12$$ for $$\sum_{r=1}^n\frac r{n^2+n+r}\leq\frac 1{n^2+n}\sum_{r=1}^nr=\frac 12$$ and on the other hand: \begin{align} \sum_{r=1}^n\frac r{n^2+n+r} &=n\sum_{r=1}^n\frac{r/n}{n+1+r/n}\frac 1n\\ &\geq n\int_0^1\frac x{n+1+x}\mathrm dx\\ &=n\left(1-(n+1)\log\left(1+\frac 1{n+1}\right)\right)\\ &\xrightarrow{n\to\infty}\frac 12 \end{align}