Calculate $\lim_{n \rightarrow \infty}n^x (a_1 a_2\dots a_n)^{\frac{1}{n}}$.

Question

Suppose that $\{a_n\}$ is a sequence such that $\displaystyle\lim_{n \rightarrow\infty} {n^x}a_n=a $ for some real $\,x$. Calculate
$$\lim_{n \rightarrow \infty}n^x (a_1\,a_2\dots\,a_n)^{\frac{1}{n}}$$

My attempts : I take $a_1=a_2 = \dots =a_n = a$

after that $\lim_{n \rightarrow \infty}$ $n^x (a_1\,a_2 \dots \,a_n)^{\frac{1}{n}}= \infty \, a = \infty$

Is it correct ?? or not

Please help me.

Any hints/soluion.....

Answer 1

Assume that $a_n> 0$ for every positive integer $n$; otherwise the limit may not exist. Set $z_n:=n^x\,a_n$ like Martin R recommended. Thus, $$n^x\,\left(\prod_{i=1}^n\,a_i\right)^{\frac{1}{n}}=\frac{n^x}{\left(\prod\limits_{i=1}^n\,i^x\right)^{\frac1n}}\,\left(\prod\limits_{i=1}^n\,z_i\right)^{\frac1n}=\left(\frac{n^n}{n!}\right)^{\frac{x}{n}}\,\left(\prod_{i=1}^n\,z_i\right)^{\frac1n}\,.$$ Now, since $\displaystyle\lim_{n\to\infty}\,z_n=a$, we have $\displaystyle\lim_{n\to\infty}\,\left(\prod_{i=1}^n\,z_i\right)^{\frac1n}=a$ (see Martin R's link in the comments above). Furthermore, Stirling's approximation $n!\approx \sqrt{2\pi n}\left(\frac{n}{\text{e}}\right)^n$ implies that $$\lim_{n\to\infty}\,\left(\frac{n^n}{n!}\right)^{\frac{x}{n}}=\lim_{n\to\infty}\,\left(\frac{\text{e}^n}{\sqrt{2\pi n}}\right)^{\frac{x}{n}}=\exp(x)\,.$$ Consequently, $$\lim_{n\to\infty}\,n^x\,\left(\prod_{i=1}^n\,a_i\right)^{\frac{1}{n}}=a\,\exp(x)\,.$$

Answer 2

Since $a_n \approx an^{-x}$,

$\begin{array}\\ n^x (\prod_{k=1}^na_k)^{\frac{1}{n}} &\approx n^x \left(\prod_{k=1}^n(ak^{-x})\right)^{\frac{1}{n}}\\ &= n^x \left(a^nn!^{-x}\right)^{\frac{1}{n}}\\ &= n^x a\left(n!^{1/n}\right)^{-x}\\ &\approx n^x a\left(n/e\right)^{-x}\\ &= ae^{x}\\ \end{array} $

Answer 3

Consider $b_n=n^{nx} a_1 a_2\dots a_n$ and then $b_{n+1}/b_n=(1+n^{-1})^{nx}(n+1)^xa_{n+1}\to e^xa$ and hence $b_n^{1/n}\to ae^x$.