Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$ Evaluate the limit 
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$ 
My approach : 
If I divide numerator and denominator by $n^2$ I get : 
$$\lim_{ n \to \infty} \dfrac{\frac{1}{n^2}}{\frac{1}{n^2} +1} +\dfrac{\frac{2}{n^2}}{\frac{2}{n^2} +1} + \ldots+ \dfrac{\frac{1}{n}}{\frac{1}{n} + 1}=0$$ 
but the answer is $\dfrac{1}{2}$ please suggest how to solve this.. thanks.
 A: Since
$$
1+n^2 \le k+n^2\le n+n^2 \quad \forall k \in \{1,2,\ldots,n\},
$$
it follows that
$$
\frac{n(n+1)}{2(n+n^2)}=\sum_{k=1}^n\frac{k}{n+n^2}\le \sum_{k=1}^n\frac{k}{k+n^2}\le\sum_{k=1}^n\frac{k}{1+n^2}=\frac{n(n+1)}{2(1+n^2)} \quad \forall n \ge 1.
$$
Thus
$$
\lim_n\sum_{k=1}^n\frac{k}{k+n^2}=\frac12.
$$
A: For each $1\leq i\leq n$, $\frac{1}{n^2+i}\leq \frac{1}{n^2}$ and $\frac{1}{n^2+i}\geq \frac{1}{n^2+n}$, and so we may bound the sum from above and below by $$\sum_{i=1}^{n}\frac{i}{n+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\sum_{i=1}^{n}\frac{i}{n^{2}}.$$ Since $\sum_{i=1}^{n}i=\frac{n(n+1)}{2},$ this becomes $$\frac{1}{2}=\frac{n(n+1)}{2n(n+1)}\leq\sum_{i=1}^{n}\frac{i}{i+n^{2}}\leq\frac{n(n+1)}{2n^{2}}=\frac{1}{2}\left(1+\frac{1}{n}\right),$$ and so it follows from the squeeze theorem that the limit is $\frac{1}{2}$.
A: $$ S_n=\sum_{k=1}^n \frac{k}{k+n^2} = \sum_{k=1}^n \left( 1 -\frac{n^2}{k+n^2} \right) \\
= n - \sum_{k=1}^n \left(\frac{n^2}{k+n^2} \right)
$$
but
$$\sum_{k=1}^n \left(\frac{n^2}{k+n^2} \right) = \sum_{k=1}^n \frac1{1+\frac{k}{n^2}}$$ 
$$ =  \sum_{k=1}^n \sum_{j=0}^{\infty} \left(\frac{-k}{n^2} \right)^j \\
= n  -\frac1{n^2}\frac{n(n+1)}{2} +O\left(\frac1{n}\right) = n - \frac12 +O\left(\frac1{n}\right) $$
so 
$$ \lim_{n \rightarrow \infty} S_n = n - (n - \frac12) = \frac12 $$
A: Use Riemann sums to show that $S\in\Big[\tfrac{\ln2}2,\tfrac12\Big]$ :
$$\sum_{k=1}^n\frac k{k^2+n^2}<\sum_{k=1}^n\frac k{k+n^2}<\sum_{k=1}^n\frac k{n^2}\quad\iff\quad\int_0^1\frac x{1+x^2}dx<S<\int_0^1xdx$$
A: Answer
$$\lim_{n \to \infty} \dfrac{1}{1+n^2} +\dfrac{2}{2+n^2}+ \ldots+\dfrac{n}{n+n^2}$$
$$\lim_{n \to \infty} \dfrac{1}{n^2}(\dfrac{1}{1/n^2 + 1} +\dfrac{2}{2/n^2+1}+ \ldots+\dfrac{n}{n/n^2+1})$$
$$\lim_{n \to \infty} \dfrac{1}{n^2}\dfrac{n*(n+1)}{2}$$
$$\lim_{n \to \infty}\dfrac{1}{2}*(1+\dfrac{1}{n})$$
$$=\dfrac {1}{2}$$
