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Compute $\displaystyle \lim_{n \rightarrow \infty} \prod_{k=1}^n \left(1+\ln\left(\frac{k+\sqrt{k^2+n^2}}{n}\right)^{\frac{1}{n}}\right)$

Note that $\frac{k+\sqrt{k^2+n^2}}{n}\geq 1$ so we're dealing with positive terms.

Taking $\log$ of the product, we're interested in the limit of $\displaystyle \sum_{k=1}^n \ln\left( 1+ \exp \left(\frac 1n\ln\ln \left( \frac kn + \sqrt{\left( \frac{k}{n} \right) ^2 +1}\right)\right) \right)$

which look very much like a Riemann sum.

Setting $f(x) =\ln\ln(x+\sqrt{x^2+1})$, the sum rewrite as $\displaystyle \sum_{k=1}^n \ln\left( 1+ \exp \left(\frac 1nf \left(\frac kn\right)\right) \right)$

I've been trying to use the usual bounds on $\ln$ and $\exp$ to exhibit a Riemann sum, but it's quite messy.

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  • $\begingroup$ You are on the right track: just perform a Taylor series for the function $\ln(1+e^y)$ with $y= f(k/n)/n$. $\endgroup$ – Fabian May 7 '16 at 18:06
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Call $$a(n,k)=\ln \left( \frac{k}{n}+\sqrt{\frac{k^2}{n^2}+1 } \right)$$ and notice that $|a(n,k)|\le C$ for some fixed constant $C$. We want to evaulate $$S_n=\sum_{k=1}^n \ln(1+1/n\ln(\frac{1}{n}a(n,k))).$$ Using $\log(1+x)=x+O(x^2)$, we can write $$S_n=\sum_{k=1}^n \frac{1}{n}a(n,k)+O(\frac{1}{n}a(n,k))=\sum_{k=1}^n \frac{1}{n}a(n,k)+O(\frac{1}{n^2})=R_n+O(1/n)$$ where $R_n$ is the n-th Riemann sum of the integral $$\int_0^1 \ln(x+\sqrt{x^2+1}) dx=I.$$ Hence $$\lim_{n \to \infty} S_n=I$$ and then the original product is $e^I$.

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  • $\begingroup$ I'm confused, we actually want to evaluate $\sum \ln(1+\exp(\frac 1n \ln (a(n,k))))$ $\endgroup$ – Gabriel Romon May 7 '16 at 18:56
  • $\begingroup$ Wait, but the exp $1/n$ is on the argument or on the log? $\endgroup$ – mrprottolo May 7 '16 at 19:04
  • $\begingroup$ It's on the $\log$ I think, otherwise there would be extra parentheses $\endgroup$ – Gabriel Romon May 7 '16 at 19:06
  • $\begingroup$ Oh, then I misread , but you can use a similar argument using taylor series and deleting the terms that go to zero. I'll change my answer later. $\endgroup$ – mrprottolo May 7 '16 at 19:10
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    $\begingroup$ @LeGrandDODOM if the exponent $1/n$ is on the whole log then your product diverges to $+ \infty$ $\endgroup$ – mrprottolo May 7 '16 at 23:19

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