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.
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} $
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.
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\lim_{n\ \to\ \infty}\sum_{k\ =\ 1}^{n}{n + k \over n^{2} + k} ={3 \over 2}:\ {\large ?}}$.
\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}
As a slightly different alternative
$$\left(\sum_{i=1}^n\frac{n + i}{n^2 + i}\right)-1=\sum_{i=1}^n\left(\frac{n + i}{n^2 + i}-\frac1n\right)=\frac{n-1}{n}\sum_{i=1}^n\frac{ i}{n^2 + i}$$
with
$$\frac12=\frac1{n^2+n}\sum_{i=1}^ni\le \sum_{i=1}^n\frac{ i}{n^2 + i}\le \frac1{n^2+1}\sum_{i=1}^ni \to \frac12$$
