3
$\begingroup$

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*}

$\endgroup$
8
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ good solution! thanks a lot. $\endgroup$ – mengdie1982 Oct 20 '19 at 16:21
3
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ 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$... $\endgroup$ – mengdie1982 Oct 20 '19 at 13:21
0
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Still_a_kid Oct 20 '19 at 13:41
  • $\begingroup$ sir, here $n=1,2,\cdots$ $\endgroup$ – mengdie1982 Oct 20 '19 at 13:47
  • $\begingroup$ 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} $. $\endgroup$ – Still_a_kid Oct 20 '19 at 15:23
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.