Does $a_{n} = \frac{x^{n}}{n^{x}}$ $a_n=x^n/n^x$ converge for any x?

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Does $a_{n} = \frac{x^{n}}{n^{x}}$ $a_n=x^n/n^x$ converge for any x?

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Answer 1 · 2018-05-27T12:11:01

$No$ $"No"$

Explanation:

$If x = - 1, we have a_{n} = n \cdot {(- 1)}^{n} and this alternates$ $"If "x = -1", we have "a_n = n*(-1)^n" and this alternates"$
$between - \infty and + \infty for n \to \infty, depending on the$ $"between "-oo" and "+oo" for "n->oo, " depending on the"$
$fact if n is odd or even.$ $"fact if n is odd or even."$
$If x < - 1, the situation gets even worse.$ $"If "x < -1", the situation gets even worse."$
$There is only convergence for x > - 1 .$ $"There is only convergence for "x > -1.$

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Does $a_{n} = \frac{x^{n}}{n^{x}}$ $a_n=x^n/n^x$ converge for any x?

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Does $a_{n} = \frac{x^{n}}{n^{x}}$ $a_n=x^n/n^x$ converge for any x?