Evaluate \(\displaystyle\lim_{n\to \infty}\left(\frac{1}{1+n^2}+\frac{2}{2+n^2}+\ldots+\frac{n}{n+n^2}\right)\)
$\frac{1}{2}$
Let \(S_n=\displaystyle\sum_{k=1}^{n}\frac{k}{k+n^2}\). Since \(1\le k\le n\), every denominator satisfies \(n^2\le k+n^2\le n^2+n\).
So \(\dfrac{k}{n^2+n}\le \dfrac{k}{k+n^2}\le \dfrac{k}{n^2}\), and summing over \(k=1\) to \(n\) (using \(\sum k=\tfrac{n(n+1)}{2}\)) gives \(\dfrac{n(n+1)/2}{n^2+n}\le S_n\le \dfrac{n(n+1)/2}{n^2}\).
The lower bound simplifies exactly: \(\dfrac{n(n+1)/2}{n(n+1)}=\dfrac12\) for every \(n\).
The upper bound is \(\dfrac{n+1}{2n}\to \dfrac12\) as \(n\to\infty\).
By the Squeeze theorem, since both bounds tend to \(\tfrac12\), \(\displaystyle\lim_{n\to\infty}S_n=\dfrac12\).
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