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

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.

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  • $\begingroup$ Here's a link to something similar: math.stackexchange.com/a/266325 $\endgroup$ – Oliver Jones Sep 26 '19 at 7:45
  • $\begingroup$ @OliverJones I think they are different,for here is only $n^2$ without $kn$ $\endgroup$ – nevermind_15 Sep 26 '19 at 7:51
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    $\begingroup$ I tried,but..$\lim _{n \rightarrow \infty} e^{n \ln \sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}}=e^{\lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n^{2}}}\right)}=e^{\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{1}{\sqrt{1+x/n}} d x}=?$ $\endgroup$ – nevermind_15 Sep 26 '19 at 8:13
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    $\begingroup$ thank everyone I got it. $\endgroup$ – nevermind_15 Sep 26 '19 at 8:27
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    $\begingroup$ @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. $\endgroup$ – Claude Leibovici Sep 27 '19 at 3:08
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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 $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 0.767294964861868504330682294138 & 0.767265822508827518149817737282 \\ 2 & 0.777590536288115341510055616562 & 0.777590505099016611262737175709 \\ 3 & 0.778679161743452556668222143405 & 0.778679161712052033718788130663 \\ 4 & 0.778788614972113403498441088036 & 0.778788614972081981769783872820 \\ 5 & 0.778799566201810759434607005418 & 0.778799566201810728010757149700 \\ 6 & 0.778800661383848807740755037386 & 0.778800661383848807709330975404 \\ 7 & 0.778800770902643295698495154985 & 0.778800770902643295698463730901 \\ 8 & 0.778800781854528651325540419889 & 0.778800781854528651325540388465 \\ 9 & 0.778800782949717245956557658910 & 0.778800782949717245956557658879 \\ 10 & 0.778800783059236106010342509938 & 0.778800783059236106010342509938 \end{array} \right)$$

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  • $\begingroup$ This solution is correct and fantastic !!!. $\endgroup$ – Dr Zafar Ahmed DSc Sep 26 '19 at 8:51
  • $\begingroup$ @DrZafarAhmedDSc. Thank you ! I felt in love with Taylor (series !) 62 years ago and I use them almost everyday. $\endgroup$ – Claude Leibovici Sep 26 '19 at 8:53
  • $\begingroup$ 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! $\endgroup$ – Sil Sep 26 '19 at 8:56
  • $\begingroup$ @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. $\endgroup$ – Dr Zafar Ahmed DSc Sep 26 '19 at 9:00
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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

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

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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 $$\ln L= n \lim_{n \rightarrow \infty} \ln S_n = \lim_{n \rightarrow \infty} n \ln \left(1-\frac{1}{4n}\right) =\lim_{n \rightarrow \infty} n \frac{-1}{4n}=\frac{-1}{4} \Rightarrow L = e^{-\frac{1}{4}}$$

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  • $\begingroup$ Another nice proof! $\endgroup$ – Oliver Jones Sep 28 '19 at 6:30
  • $\begingroup$ @Oliver Jones Thanks $\endgroup$ – Dr Zafar Ahmed DSc Sep 28 '19 at 8:13
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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

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

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