How to prove $ \lim_{n\to\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$? How to prove that
$\displaystyle \lim_{n\longrightarrow\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$?
I suppose some bounds are nedded, but the ones I have found are not sharp enough (changing $k$ for $1$ or $n$ leads to the limit being between 1 and 2).
Any suggestion is welcomed.
 A: Rewrite the expression as $\displaystyle\sum_{k=1}^{n}\dfrac{n+k}{n^2+k} = \dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}}$. 
For all $1 \le k \le n$ we have $\dfrac{n}{n+1}\left(1 + \dfrac{k}{n}\right) = \dfrac{1 + \frac{k}{n}}{1+\frac{1}{n}} \le \dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}} \le \dfrac{1 + \frac{k}{n}}{1+0} = 1 + \dfrac{k}{n}$.
Therefore, $\displaystyle\dfrac{n}{n+1} \cdot \dfrac{1}{n}\sum_{k = 1}^{n}\left(1+\dfrac{k}{n}\right) \le \dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1+\frac{k}{n}}{1+\frac{k}{n^2}} \le \dfrac{1}{n}\sum_{k = 1}^{n}\left(1+\dfrac{k}{n}\right)$. 
Now, use the fact that $\displaystyle\lim_{n \to \infty}\dfrac{n}{n+1} = 1$ and that $\displaystyle\dfrac{1}{n}\sum_{k = 1}^{n}\left(1+\dfrac{k}{n}\right)$ is a Riemann sum for $\displaystyle\int_{0}^{1}(1+x)\,dx$ to get the result.  
A: Squeeze (without integral calculus).
I. $\displaystyle\sum_{k=1}^{n}\dfrac{n+k}{n^2+k} \leq \sum_{k=1}^{n}\dfrac{n+k}{n^2+1}=\frac{1}{n^2+1}\sum_{k=1}^{n}(n+k) = \frac{1}{n^2+1}(n^2+\frac{n(n+1)}{2})\rightarrow\frac{3}{2} $.
II.$\displaystyle\sum_{k=1}^{n}\dfrac{n+k}{n^2+k} \geq\sum_{k=1}^{n}\dfrac{n+k}{n^2+n}=\frac{1}{n^2+n}\sum_{k=1}^{n}(n+k) = \frac{1}{n^2+n}(n^2+\frac{n(n+1)}{2})\rightarrow\frac{3}{2} $
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$\ds{\lim_{n\ \to\ \infty}\sum_{k\ =\ 1}^{n}{n + k \over n^{2} + k}
     ={3 \over 2}:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large%
\lim_{n\ \to\ \infty}\sum_{k\ =\ 1}^{n}{n+k \over n^{2} + k}}
=\lim_{n\ \to\ \infty}\bracks{%
n + \pars{n - n^{2}}\sum_{k\ =\ 1}^{n}{1 \over k + n^{2}}}
\\[5mm]&=\lim_{n\ \to\ \infty}\bracks{%
n + \pars{n - n^{2}}\sum_{k\ =\ 1}^{n}\int_{0}^{1}t^{k - 1 + n^{2}}\,\dd t}
\\[5mm]&=\lim_{n\ \to\ \infty}\bracks{%
n + \pars{n - n^{2}}\int_{0}^{1}t^{n^{2}}\sum_{k\ =\ 1}^{n}t^{k - 1}\,\dd t}
\\[5mm]&=\lim_{n\ \to\ \infty}\bracks{%
n + \pars{n - n^{2}}\int_{0}^{1}t^{n^{2}}\,{1 - t^{n} \over 1 - t}\,\dd t}
\\[5mm]&=\lim_{n\ \to\ \infty}\bracks{%
n + \pars{n - n^{2}}\pars{%
\int_{0}^{1}{1 - t^{n^{2} + n} \over 1 - t}\,\dd t
-\int_{0}^{1}{1 - t^{n^{2}} \over 1 - t}\,\dd t}}
\\[5mm]&=\lim_{n\ \to\ \infty}\bracks{%
n + \pars{n - n^{2}}\pars{H_{n^{2} + n} - H_{n^{2}}}}
\end{align}
  where $\ds{H_{m}}$ is a Harmonic Number.

However, when $\ds{n \ggg 1}$:
$$
H_{n^{2} + n} - H_{n^{2}}\sim\ln\pars{1 + {n \over n^{2} + 1}}
\sim {n \over n^{2} + 1} - {n^{2} \over 2\pars{n^{2} + 1}^{2}}
={2n^{3} - n^{2} + 2n \over 2\pars{n^{2} + 1}^{2}}
$$

Then,
  \begin{align}
&\color{#66f}{\large%
\lim_{n\ \to\ \infty}\sum_{k\ =\ 1}^{n}{n+k \over n^{2} + k}}
=\lim_{n\ \to\ \infty}\bracks{%
n + \pars{n - n^{2}}\,{2n^{3} - n^{2} + 2n \over 2\pars{n^{2} + 1}^{2}}}
\\[5mm]&=\lim_{n\ \to\ \infty}\bracks{%
{\dsc{3}n^{4} + n^{3} +2n^{2} + 2n \over \dsc{2}\pars{n^{2} + 1}^{2}}}
=\color{#66f}{\large{3 \over 2}}
\end{align}

