Evaluate $\lim\limits_{n \to \infty}\sum\limits_{k=0}^n \dfrac{\sqrt{n}}{n+k^2}(n=1,2,\cdots)$ I tried to change it into a Riemann sum but failed, since 
\begin{align*}
\lim_{n \to \infty}\sum_{k=0}^n \frac{\sqrt{n}}{n+k^2}=\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^n \frac{\sqrt{n}}{1+(k/\sqrt{n})^2}
,\end{align*}
which is not a standard form. Maybe, it need apply the squeeze theorem, but how to evaluate the bound.
By the way, WA gives its result 
\begin{align*}
\lim_{n \to \infty}\sum_{k=0}^n \frac{\sqrt{n}}{n+k^2}=\frac{\pi}{2}.
\end{align*}
 A: Integral bounds do the job:  $$\int_k^{k+1} \frac{1}{n+t^2} dt \leq \frac{1}{n+k^2}\leq \int_{k-1}^{k} \frac{1}{n+t^2} dt$$
Summing the left-hand side inequalities for $k\in \{0,\ldots,n\}$ and the right-hand side inequalities for $k\in \{1,\ldots,n\}$ yields
$$\int_0^{n+1}\frac{1}{n+t^2} dt \leq  \sum_{k=0}^n \frac{1}{n+k^2} \leq \frac 1n + \int_{0}^{n} \frac{1}{n+t^2}$$
that is 
$$\frac{1}{\sqrt{n}}\arctan\left(\frac{n+1}{\sqrt n}\right)\leq \sum_{k=0}^n \frac{1}{n+k^2} \leq \frac 1n + \frac{1}{\sqrt{n}}\arctan\left(\frac{n}{\sqrt n}\right)$$
Multiplying by $\sqrt n$ and squeezing yields 
$$\lim_n \sqrt n \sum_{k=0}^n \frac{1}{n+k^2} = \lim_n \arctan\left(\frac{n}{\sqrt n}\right) = \frac \pi 2$$
A: $$\lim_{n\to\infty} \sum_{k=0}^n \frac{\sqrt{n}}{n+k^2}=\lim_{n\to\infty} \frac{1}{n}\sum_{k=0}^n\frac{\sqrt{n}}{1+(k/\sqrt{n})^2}$$
Now make the substitution $n=m^2$, so that we have
$$\lim_{n\to\infty} \frac{1}{m}\sum_{k=0}^n\frac{1}{1+(k/m)^2} $$
which is equal to
$$\lim_{x\to\infty }\lim_{m\to\infty} \sum_{k=0}^{mx}\frac{x}{mx}\frac{1}{1+(kx/mx)^2} =\lim_{x\to\infty }\int_0^x \frac{dk}{1+k^2}=\int_0^\infty \frac{dk}{1+k^2}$$
which is, of course, equal to $\frac{\pi}{2}$.
A: you could let: $n\rightarrow n^2$ then we get:
$$L=\lim_{n\to\infty}\sum_{k=1}^\infty\frac{n}{n^2+k^2}$$
and we know that:
$$\frac{1}{n^2+k^2}\le\frac{n}{n^2+k^2}\le\frac{n^2}{n^2+k^2}$$
Now from here we can see that:
$$\sum_{k=-\infty}^\infty\frac{1}{k^2+a^2}=\frac\pi a\coth(\pi a)$$
and notice that:
$$\sum_{k=-\infty}^\infty=\sum_{k=-\infty}^{-1}+f(0)+\sum_{k=1}^\infty,\,\,\sum_{k=-\infty}^{-1}=\sum_{k=1}^\infty$$
and so:
$$\sum_{k=1}^\infty\frac{1}{k^2+a^2}=\frac12\left[\frac\pi a\coth(\pi a)-\frac1{a^2+1}\right]$$
This gives us:
$$\frac12\left[\frac\pi n\coth(\pi n)-\frac1{n^2+1}\right]\le\sum_{k=1}^\infty\frac{n}{n^2+k^2}\le\frac12\left[\pi n\coth(\pi n)-\frac{n^2}{n^2+1}\right]$$
Hope this is correct!, Although I don't think the bounds are particularly tight
A: We make a substitution $t^2=k $,
$$\lim_{n \to \infty}\sum_{t=0}^\sqrt {n} \frac{\sqrt{n}}{n+t}=\lim_{n \to \infty}\sum_{t=0}^\sqrt {n} \frac{1}{1+\frac{t}{\sqrt{n}}} \frac {1} {\sqrt{n}} $$
In a normal Riemann Transformation I would replace $\frac {1}{n}$ with $dx $ (where $x= \frac {t}{n}$) but since we have $\frac {1}{\sqrt {n}}$ we would replace it with $2dx $ (where $x= \frac {t}{\sqrt {n}}$) 
Hence the limit becomes 
$$2 \int_ 0^1 \frac {1}{1+x^2}\,dx = \frac {\pi}{2}$$
