<s>[INST] 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.
 [/INST] 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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