Finding limit of a product.

Question

Prove:$$\lim_{n \to\infty }\frac{1}{n}\left[\prod_{i=1}^{n}(n+i) \right ]^{\frac{1}{n}}=\frac{4}{e}$$ I tried using Squeeze Theorem but can't go beyond $1<L<2$. $$\lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n=e$$might be useful too.

Answer 1

$\frac{1}{n}\left[\prod_{i=1}^n(n+i)\right]^{1/n}=\left[\prod_{i=1}^n \frac{1}{n}(n+i)\right]^{1/n}=\left[\prod_{i=1}^n (1+\frac{i}{n})\right]^{1/n}$ Taking log we get $\frac{1}{n}\sum_{i=1}^n\ln (1+\frac{i}{n}) \to \int_0^1 \ln(1+x)dx, n \to \infty$ Integrating by parts gives $\int_0^1 \ln(1+x)dx=\ln 4 -1.$ Now the limit of the product is $e^{\ln 4 - 1}$.

Answer 2

Wo consider the log. of $\frac{1}{n}\left[\prod_{i=1}^{n}(n+i) \right ]^{\frac{1}{n}}$. If you take logarithm then you get $$-\log n+\frac{1}{n}\sum_{k=1}^\infty \log(n+k)=-\log n +\frac{1}{n}\log \frac{(2n)!}{n!}$$ By Stirling's formula, $$ \begin{aligned} -\log n +\frac{1}{n}\log \frac{(2n)!}{n!}&\sim -\log n +\frac{1}{n} \log \frac{\sqrt{4\pi n}2^{2n} n^{2n} e^{-2n}}{\sqrt{2\pi n}n^n e^{-n}} \\ &= -\log n +\frac{1}{n} \log \left( \sqrt{2}\cdot 4^n n^ne^{-n}\right)\\ &= -\log n +\frac{1}{n}\left( \log\sqrt{2} +n\log 4+n\log n-n\right)\\ &= \frac{1}{n}\log \sqrt{2} +\log 4 -1 \end{aligned}$$

So $\lim_{n\to\infty}-\log n+\frac{1}{n}\sum_{k=1}^\infty \log(n+k)=\log 4 -1$. It follows desired formula.

Answer 3

Note that $$\ln\left\{\frac 1n\left[\prod_{i=1}^{n}(n+i) \right ]^{\frac{1}{n}}\right\}=\ln(1/n)+(1/n)S$$ where $$S=\sum_{k=n+1}^{2n+1}\ln k.$$

Here, you can use $$\int_{n}^{2n}\ln xdx\lt S\lt \int_{n+1}^{2n+1}\ln xdx$$ $$\iff f(2n)-f(n)\lt S\lt f(2n+1)-f(n+1)$$ where $$f(n)=n(\ln(n)-1).$$

Now you'll see $$\lim_{n\to\infty}\left(\ln\frac 1n+\frac 1n(f(2n)-f(n))\right)=-1+\ln 4,$$ $$\lim_{n\to\infty}\left(\ln\frac 1n+\frac 1n(f(2n+1)-f(n+1))\right)=-1+\ln 4.$$

Hence, by Squeeze Theorem, you'll get the answer you wrote.