# 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*}

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 $$\lim_n \sqrt n \sum_{k=0}^n \frac{1}{n+k^2} = \lim_n \arctan\left(\frac{n}{\sqrt n}\right) = \frac \pi 2$$

• good solution! thanks a lot. – mengdie1982 Oct 20 '19 at 16:21

$$\lim_{n\to\infty} \sum_{k=0}^n \frac{\sqrt{n}}{n+k^2}=\lim_{n\to\infty} \frac{1}{n}\sum_{k=0}^n\frac{\sqrt{n}}{1+(k/\sqrt{n})^2}$$ 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 $$\frac{\pi}{2}$$.

• THX！ But notice that，if you substitute $m^2$ for $n$，you just prove a certain subsequence converges only. Maybe you should consider squeeze that $m^2 \leq n<(m+1)^2$... – mengdie1982 Oct 20 '19 at 13:21

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

$$2 \int_ 0^1 \frac {1}{1+x^2}\,dx = \frac {\pi}{2}$$

• I am new to MathJax (and MSE). I would explain why we take it as $2dx$ in the integral, but that would make a lot of syntax errors creep in. But I think the reason is obvious. – Still_a_kid Oct 20 '19 at 13:41
• sir, here $n=1,2,\cdots$ – mengdie1982 Oct 20 '19 at 13:47
• The substitution $r=t^2$ would make each iteration go like $\sqrt {1}, \sqrt {2}, \sqrt {3}, \cdots$ instead of $1, 2, 3, \cdots$. Which would enable $t$ to achieve $\sqrt {n}$. – Still_a_kid Oct 20 '19 at 15:23

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