# Limit of the sequence $x_n= \frac{[(a+1)(a+2)…(a+n)]^{1/n}}{n}$

Limit of the sequence $$x_n= \frac{[(a+1)(a+2)...(a+n)]^{1/n}}{n}.$$ where $$a$$ is a fixed positive real number.

I want to find the limit of the sequence . I tried to apply Cauchy limit theorem. But I failed. How to find the limit of this sequence

• Any conditions on $a$? – rtybase Feb 3 at 16:28
• sorry, a is fixed positive real number – Cloud JR Feb 3 at 16:33
• Intuitively, taking $a$ to be integer, the product is the factorial of $n+a$ over $a!$. Taking the $n^{th}$ root, $a!$ vanishes and using the Stirling approximation we are left with $(a+n)/ne$ which converges to $1/e$. – Yves Daoust Feb 3 at 18:31

Take logs and apply Stolz Cesaro. Indeed $$x_n=\frac{1}{n}\prod_{k=1}^n(a+k)^{1/n}$$ so $$\log x_n=\frac{1}{n^2}\sum_{k=1}^n\log(a+k)$$ whence $$\lim_{n\to\infty} \log x_n=\lim_{n\to \infty}\frac{\log(a+n+1)}{2n+1}$$ a limit that you should be able to compute.

• Shouldn't the expression for the logarithm be $\log x_n=-\log n+\frac{1}{n}\sum\limits_{k=1}^n\log(a+k) =\frac{-n\log n+\sum\limits_{k=1}^n\log(a+k)}n$. (Still, Stolz-Cesaro theorem can help here - similar as in the linked posts.) – Martin Sleziak Feb 3 at 17:41
• Your answer is not correct (you get $\log x_n \to 0$, $x_n \to 1$). Why don't you fix it (as Martin pointed out) so that future readers are not misled by a wrong accepted answer? – Martin R Feb 4 at 5:59

One can proceed similarly as in Evaluate $\lim_{n \rightarrow \infty} \frac {[(n+1)(n+2)\cdots(n+n)]^{1/n}}{n}$: Use that $$\lim\limits_{n \to \infty} \sqrt[n]{a_n} = \lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n}$$ if all $$a_n$$ are positive and if the limit on the right-hand side exists. (See for example Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$.)

Here $$a_n = \frac{(a+1)\ldots (a+n)}{n^n}$$ and $$\frac{a_{n+1}}{a_n} = \frac{(a+n+1)n^n}{(n+1)^{n+1}} = \frac{a+n+1}{n+1} \left( \frac{n}{n+1} \right)^n \to \frac 1e \, .$$

In terms of the Rising Factorial and Gamma function you have: $$\prod\limits_{k = 1}^n {\left( {a + k} \right)} = \prod\limits_{k = 0}^{n - 1} {\left( {a + 1 + k} \right)} = \left( {a + 1} \right)^{\,\overline {\,n\,} } = {{\Gamma \left( {a + 1 + n} \right)} \over {\Gamma \left( {a + 1} \right)}}$$

By the Stirling approximation we have $$z^{\,\overline {\,w\,} } \propto {{\sqrt {2\,\pi } } \over {\Gamma (z)}}e^{\, - \,w} w^{\,z + w - 1/2} \left( {1 + O\left( {{1 \over w}} \right)} \right) \quad \left| \matrix{ \;\left| w \right| \to \infty \hfill \cr \;\left| {\arg (z + w)} \right| < \pi \hfill \cr} \right.$$

So \eqalign{ & {1 \over n}\left( {\left( {a + 1} \right)^{\,\overline {\,n\,} } } \right)^{\,1/n} \propto \left( {{{\sqrt {2\,\pi } } \over {\Gamma (a + 1)}}} \right)^{\,1/n} e^{\, - \,1} \;n^{\,{{a + 1/2} \over n}} \left( {1 + O\left( {{1 \over n}} \right)} \right)^{\,1/n} = \cr & = \left( {{{\sqrt {2\,\pi } } \over {\Gamma (a + 1)}}} \right)^{\,1/n} e^{\, - \,1 + \left( {{{a + 1/2} \over n}} \right)\;\ln n} \left( {1 + O\left( {{1 \over n}} \right)} \right)^{\,1/n} \cr}

and $$\mathop {\lim }\limits_{n\; \to \;\infty } {1 \over n}\left( {\left( {a + 1} \right)^{\,\overline {\,n\,} } } \right)^{1/n} = 1/e$$