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

My attempts : i take $a_1=a_2 =.......=a_n = a$

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

Is it correct ?? or not

pliz help me,

any HINTS/SOLUTion.....

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  • $\begingroup$ Or this one: math.stackexchange.com/q/352935/42969. $\endgroup$ – Martin R Jul 20 '18 at 11:29
  • $\begingroup$ Well, the first error that you make is in taking $a_i = a$. Since, then $\lim n^x a_n = a \lim n^x $ which equals $0$ if $x<0$, $a$ if $x=0$ and $\text{sign}(a)\infty$ if $x>0$. $\endgroup$ – Stan Tendijck Jul 20 '18 at 11:30
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    $\begingroup$ No, there will be an extra factor. Note $$n^x\,\left(\prod_{i=1}^n\,a_i\right)^{\frac{1}{n}}=\frac{n^x}{\left(\prod_{i=1}^n\,i^x\right)^{\frac1n}}\,\left(\prod_{i=1}^n\,x_i\right)^{\frac1n}=\left(\frac{n^n}{n!}\right)^{\frac{x}{n}}\,\left(\prod_{i=1}^n\,x_i\right)^{\frac1n}\,.$$ $\endgroup$ – Batominovski Jul 20 '18 at 11:44
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    $\begingroup$ @MartinR But please don't delete both links. They are very useful to this problem. $\endgroup$ – Batominovski Jul 20 '18 at 11:47
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    $\begingroup$ @JoséCarlosSantos This is not a duplicate. Please read the comments. $\endgroup$ – Batominovski Jul 20 '18 at 13:54
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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, $$\lim_{n\to\infty}\,n^x\,\left(\prod_{i=1}^n\,a_i\right)^{\frac{1}{n}}=a\,\exp(x)\,.$$

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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} $

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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 $b_n^{1/n}\to ae^x$.

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