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.

  • $\begingroup$ 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

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 $e^I$.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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.