Evaluate $$\lim_{n\to\infty} \frac{1}{n} \left( (n+1)\cdots (n+n) \right)^\frac{1}{n}$$

I somehow need to get a Riemann sum from this expression in order to evaluate the limit. I tried to use $f(x)^{g(x)}= e^{\ln(f(x))g(x)}$ identity but got stuck.

$$ e^{ \ln(\frac{1}{n}(n+1)^\frac{1}{n}\cdots (n+n)^\frac{1}{n})} $$

Then, $$\lim_{n\to\infty} \ln(\frac{1}{n}(n+1)^\frac{1}{n}\cdots (n+n)^\frac{1}{n}) = \lim_{n\to\infty} \ln(\frac{1}{n}) + \sum_{i=1}^n \ln(i+1)^\frac{1}{n}$$

  • $\begingroup$ I would start by looking at the logarithm of the expression which basically leaves you with $1/n$ times a sum of logarithms. $\endgroup$ – Dr_Be Jan 18 '16 at 9:22
  • $\begingroup$ You can take the $\frac 1n=\left(\frac 1{n^n}\right)^{\frac 1n}$ inside the product before doing anything else, if that helps. $\endgroup$ – Mark Bennet Jan 18 '16 at 9:25
  • $\begingroup$ The product $(n+1) \dots (n+n)$ is the ratio of two $\Gamma$ functions, which can be reduced to $\Gamma(2 n)/\Gamma(n)$. Then use dlmf.nist.gov/5.5.E5 and Stirling's formula. $\endgroup$ – Johannes Trost Jan 18 '16 at 9:27

$\frac{1}{n} \left( (n+1)\cdots (n+n) \right)^\frac{1}{n}=\frac{1}{n}(n^n(1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{n}{n}))^{1/n}=((1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{n}{n}))^{1/n}$.

Then $((1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{n}{n}))^{1/n}=e^{\frac{1}{n}\sum_{k=1}^n\log(1+\frac{k}{n})}\to e^{\int_0^1\log(1+x)dx}$, which is easy to calculate.


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.