Let

$$x_n = \sum_{k=1}^{n}\frac{n+k}{n^2+k},\qquad y_n=\sum_{k=1}^{n}\frac{n+k}{n^2+n},\qquad z_n=\sum_{k=1}^{n}\frac{n+k}{n^2+1}$$

Then, it is clear that $y_n\le x_n\le z_n$, and

$$\begin{align*} y_n &= \sum_{k=1}^{n}\frac{n+k}{n^2+n} = \frac{n^2 + \frac{n(n+1)}{2}}{n^2 + n}=\frac{3n+1}{2n+2} &\implies \lim_{n\to\infty}y_n=\frac{3}{2}\\ z_n &= \sum_{k=1}^{n}\frac{n+k}{n^2+1} = \frac{n^2 + \frac{n(n+1)}{2}}{n^2 + 1}=\frac{3n^2+n}{2n^2+2} &\implies \lim_{n\to\infty}z_n=\frac{3}{2}\end{align*}$$

Therefore, by the sandwich theorem, we find

$$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{n+k}{n^2 + k}=\lim_{n\to\infty}x_n=\boxed{\frac{3}{2}}$$