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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*}

mengdie1982
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4 Answers4

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

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

Franklin Pezzuti Dyer
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    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
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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

Henry Lee
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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

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

srswat
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  • 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. – srswat 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} $. – srswat Oct 20 '19 at 15:23