9

$\displaystyle\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+1}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+n}}\right)^{n}$

left=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+1}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+1}}} }=e^{0}=1$

right=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+n}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+n}}} }=e^{-\frac{1}{2}}$

left $\ne$ right ,what to do next?

$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{\displaystyle n \ln \sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}(1)}$

$(1)=\displaystyle \lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n^{2}}}\right)$$=\lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n \cdot 1/n}}\right)$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{1}{\sqrt{1+x/n}} d x$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{nd(x/n+1)}{\sqrt{1+x/n}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 \left.\sqrt{1+\frac{x}{n} }\right|_{0}^{1}}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 (\sqrt{1+\frac{1}{n}}-1)}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} -\frac{1}{2n^2} +o(\frac{1}{n^2}))}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} +\left(\frac{1}{2!}\cdot \frac{1}{2} \cdot \left(\frac{1}{2}-1\right) \right)\frac{1}{n^2} +o(\frac{1}{n^2}))}}=\lim_{n\to\infty}{n \ln{\left(1-\frac{1}{4n}\right)}}=-\frac{1}{4} $

so that $\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{(1)}=e^{-\frac{1}{4}}$

this solution is right.

nevermind_15
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  • Here's a link to something similar: https://math.stackexchange.com/a/266325 – Oliver Jones Sep 26 '19 at 07:45
  • @OliverJones I think they are different,for here is only $n^2$ without $kn$ – nevermind_15 Sep 26 '19 at 07:51
  • That's a superficial difference. I think the method still applies. Did you try? – Oliver Jones Sep 26 '19 at 07:56
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    I tried,but..$\lim {n \rightarrow \infty} e^{n \ln \sum{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}}=e^{\lim {n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum{k=1}^{n} \frac{1}{\sqrt{1+k/n^{2}}}\right)}=e^{\lim {n \rightarrow \infty} n \ln \int{0}^{1} \frac{1}{\sqrt{1+x/n}} d x}=?$ – nevermind_15 Sep 26 '19 at 08:13
  • @nevermind_15 You are close, have you tried substitution $y=1+x/n$ in the integral? – Sil Sep 26 '19 at 08:23
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    thank everyone I got it. – nevermind_15 Sep 26 '19 at 08:27
  • Nice! Consider answering your own question then, so it will be useful for others when same problem arises in the future. – Sil Sep 26 '19 at 08:41
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    @OliverJones. For the question you gave the link of, I used a similar approach. If you have time to waste, have a look at my answer. Cheers. – Claude Leibovici Sep 27 '19 at 03:08
  • @ClaudeLeibovici I saw your proof to the question in the link; very nice! – Oliver Jones Sep 28 '19 at 02:23
  • @nevermind_15 Your answer doesn't agree with the one given by Claude Leibovici and others. The factor $n$ might prevent you from replacing the sum with an integral. – Oliver Jones Sep 28 '19 at 02:28
  • @nevermind_15 The hole in your argument can be fixed. If we let $a_n=\sum_{k=1}^n\frac{1}{n^2+k}$, then you've shown $a_n=2n(\sqrt{1+1/n}-1)$. Your sequence is $a_n^n=e^{n\ln a_n}$. Your mistake lies in the Taylor series you're using. You should get $\lim_{n\rightarrow \infty}n\ln a_n=-1/4$. – Oliver Jones Sep 28 '19 at 06:25
  • @OliverJones thank you I fixed it – nevermind_15 Sep 28 '19 at 08:30
  • @nevermind_15 Sorry but what I said was incorrect. There are some major flaws in your proof which I don't think can be fixed. Firstly, you can't use the methods in the link to find $\lim_{n\rightarrow \infty}a_n$ with $a_n=\sum_{k=1}^n\frac{1}{n^2+k}$ due to the factor of $1/n$. If you think about it, how can the answer depend on $n$? The other problem is that you're assuming $\lim_{n\rightarrow \infty}na_n= \lim_{n\rightarrow \infty} (n\lim_{n\rightarrow \infty}a_n)$ which needs explanation. Strangely, you still got the correct answer. – Oliver Jones Sep 28 '19 at 21:23
  • @OliverJones I think the solution might be correct by some theorem that making the limit and the integral can be exchanged,$\lim_{n\to \infty}\sum_{k=1}^{n}f(k/n^2)1/n=\lim_{n\to \infty}(\lim_{n\to \infty}\sum_{k=1}^{n}f(k/n \cdot 1/n)1/n)=\lim_{n\to\infty}\int_{0}^{1}f(x/n)dx$ – nevermind_15 Sep 29 '19 at 05:24
  • @OliverJones I think the theorem might be the Uniform Continuity or Lebesgue's dominated convergence theorem,but I'm not sure. – nevermind_15 Sep 29 '19 at 05:31
  • @nevermind_15 Let me think about that for a bit. By the way, there's a typo in my comment; I should have written $\ln a_n$, not $a_n$. – Oliver Jones Sep 29 '19 at 06:52
  • @nevermind_15 I don't see how the dominated convergence theorem will help here. You can fix the problem by using techniques from asymptotic analysis but that's basically what Claude Leibovici has done below. – Oliver Jones Sep 30 '19 at 03:13

5 Answers5

6

If you know the generalized harmonic numbers, $$S_n=\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}=H_{n^2+n}^{\left(\frac{1}{2}\right)}-H_{n^2}^{\left(\frac{1}{2}\right)}$$ Using the asymptotics $$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac 1 {2\sqrt p}-\frac 1{24p\sqrt p}+O\left(\frac{1}{p^{7/2}}\right)$$ apply it twice and continue with Taylor series to get $$S_n=1-\frac{1}{4 n}-\frac{1}{8 n^2}+\frac{7}{64 n^3}+O\left(\frac{1}{n^4}\right)$$ $$\log(S_n)=-\frac{1}{4 n}-\frac{5}{32 n^2}+\frac{7}{96 n^3}+O\left(\frac{1}{n^4}\right)$$ $$n\log(S_n)=-\frac{1}{4 }-\frac{5}{32 n}+\frac{7}{96 n^2}+O\left(\frac{1}{n^3}\right)$$ $$e^{n\log(S_n)}=\frac{1}{\sqrt[4]{e}}\left(1-\frac{5}{32 n}+\frac{523}{6144 n^2}+ O\left(\frac{1}{n^3}\right)\right)$$

Edit

For the fun of it, let us compute using $n=10^k$ to get $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 0.767294964861868504330682294138 & 0.767265822508827518149817737282 \\ 2 & 0.777590536288115341510055616562 & 0.777590505099016611262737175709 \\ 3 & 0.778679161743452556668222143405 & 0.778679161712052033718788130663 \\ 4 & 0.778788614972113403498441088036 & 0.778788614972081981769783872820 \\ 5 & 0.778799566201810759434607005418 & 0.778799566201810728010757149700 \\ 6 & 0.778800661383848807740755037386 & 0.778800661383848807709330975404 \\ 7 & 0.778800770902643295698495154985 & 0.778800770902643295698463730901 \\ 8 & 0.778800781854528651325540419889 & 0.778800781854528651325540388465 \\ 9 & 0.778800782949717245956557658910 & 0.778800782949717245956557658879 \\ 10 & 0.778800783059236106010342509938 & 0.778800783059236106010342509938 \end{array} \right)$$

Claude Leibovici
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  • This solution is correct and fantastic !!!. – Z Ahmed Sep 26 '19 at 08:51
  • @DrZafarAhmedDSc. Thank you ! I felt in love with Taylor (series !) 62 years ago and I use them almost everyday. – Claude Leibovici Sep 26 '19 at 08:53
  • Is there some trick to get series $\log(S_n)$ by knowing series of $S_n$? Similarly for last one, it does not look like just exponentiation... (it's probably something basic). By the way nice! – Sil Sep 26 '19 at 08:56
  • @Claude Lebovici Yes, a true love. I have never seen so interesting solution of a limit. Also up -vote to the question itself. Oh! my best wau got only $1/e$ which is of course wrong. – Z Ahmed Sep 26 '19 at 09:00
3

We have

$$\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim_{n\to\infty} \frac1{n^n}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k/n^2}}}\right)^{n}}$$

and

$$\frac{1}{\sqrt{1+k/n^2}}=1-\frac12\frac{k}{n^2}+O\left(\frac{k^2}{n^4}\right)$$

therefore

$$\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k/n^2}}}=n-\frac12\frac{n(n+1)}{2n^2}+O\left(\frac1n\right)$$

and

$$\frac1{n^n}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k/n^2}}}\right)^{n}}=\left(1-\frac{n+1}{4n^2}+O\left(\frac1{n^2}\right)\right)^n\to\frac1{\sqrt[4] e}$$

user
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3

Another solution: Let $$S_n= \sum_{k=1}^{n} \frac{1}{n}\frac{1}{\sqrt{1+\frac{k}{n^2}}}$$ We will prove that $$ \frac{2x+2}{2+x}<\sqrt{1+x} < 1+\frac{x}{2}, x>0~~~~(1)$$ Right one: is nothing but $$2\sqrt{1+x}=-1-(1+x)-[1-\sqrt{1+x}]^2 \le 0.$$
The left one isnothing but $$\frac{2(1+x)}{1+1+x} < \sqrt{1+x} \Rightarrow 2\sqrt{1+x}<1+(1+x) \Rightarrow -[1-\sqrt{1+x}]<0.$$ From (1) it follows that $$ 1+ \frac{k}{2n^2+k} ~<~\sqrt{1+\frac{k}{n^2}}~ < ~1+\frac{k}{2n^2}$$ $$\Rightarrow 1+ \frac{k}{2n^2+\underline{2n}} ~<~\sqrt{1+\frac{k}{n^2}}~ < ~1+\frac{k}{2n^2}$$ $$\Rightarrow \left(1+ \frac{k}{2n^2+2n}\right)^{-1} ~>~\frac{1}{\sqrt{1+\frac{k}{n^2}}}~ > ~\left( 1+\frac{k}{2n^2} \right)^{-1}$$ $$\Rightarrow \left(1- \frac{k}{2n^2+2n}\right) ~>~\frac{1}{\sqrt{1+\frac{k}{n^2}}}~ > ~\left( 1-\frac{k}{2n^2} \right)$$ $$\Rightarrow 1- \sum_{k=1}^{n} \frac{k}{n(2n^2+2n)} ~>~\sum_{k=1}^{n} \frac{1}{n}\frac{1}{\sqrt{1+\frac{k}{n^2}}}~ >~ 1-\sum_{k=1}^{n}\frac{k}{2n^3}$$ $$\Rightarrow 1-\frac{1}{4n}~ > ~S_n~ >~1-\frac{1}{4n}-\frac{1}{4n^2}.$$ Now $$\ln L= n \lim_{n \rightarrow \infty} \ln S_n = \lim_{n \rightarrow \infty} n \ln \left(1-\frac{1}{4n}\right) =\lim_{n \rightarrow \infty} n \frac{-1}{4n}=\frac{-1}{4} \Rightarrow L = e^{-\frac{1}{4}}$$

Z Ahmed
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3

We can use the sandwich theorem twice for solving the limit $$\text { Observe that } \frac{n}{\sqrt{n^2 +n}}\le\alpha_n=\sum_{k=1}^{n}\frac{1}{\sqrt{n^2 +k}}\le\frac{n}{\sqrt{n^2+1}}, \text{ So, } \lim_{n\to\infty}\alpha_n=1.\space \text { Proceeding as an usual case of } 1^{\infty}, $$ $$ \alpha=\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{1}{\sqrt{n^2 +k}}\right) ^n=e^{\lim_{n\to\infty} n(\alpha_n -1)}=e^{\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{n}{\sqrt{n^2+k}} -1\right)}=e^{\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{-k}{(n+\sqrt{n^2+k})\sqrt{n^2+k}} \right)}$$ $$\text{ Now, use Sandwich theorem again and obtain that } \alpha=\frac{1}{\sqrt[4]{e}}$$

Jack
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  • I didnt quite get it first sandwich tells us its between e^-1/2 and 0 but after that what you concluded ? And what you mean by again sandwich theorem? – ProblemDestroyer Apr 22 '22 at 21:17
  • @ProblemDestroyer, you are right, I have forgotten the nth power when I wrote alpha. Now it is clear hopefully. The 1st squeeze th tells me that the limit is in the form one to infinity, after that the 2nd squeeze(applied to the exponent of e) gives the solution we were looking for. – Jack Apr 23 '22 at 17:51
  • Thanks for explaining really nice method . First time seeing sandiwich theorem telling us about the form and then again applied . :) – ProblemDestroyer Apr 23 '22 at 23:51
1

Following your solution from here

$$...= \lim_{n\to\infty}{n\ln{n \cdot2 \left(\sqrt{1+\frac{1}{n}}-1\right)}}=...$$

we have that

$$\sqrt{1+\frac{1}{n}}-1=\frac1{2n}-\frac1{8n^2}+O\left(\frac1{n^3}\right)$$

and therefore

$$\lim_{n\to\infty}{n\ln{n \cdot2 \left(\sqrt{1+\frac{1}{n}}-1\right)}}=\lim_{n\to\infty}{n\ln{\left(1-\frac1{4n}+O\left(\frac1{n^2}\right)\right)}}=-\frac14$$

user
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