For what values of $r$ $r$ does the sequence $a_{n} = {(n r)}^{n}$ $a_n = (nr)^n$ converge ?

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Answer 1 · 2017-09-17T21:40:01

$r = 0$ $r=0$

If $r \neq 0$ $r != 0$ then when $n \geq ∣ ∣ ∣ \frac{2}{r} ∣ ∣ ∣$ $n >= abs(2/r)$ we have:

$∣ ∣ {(n r)}^{n} ∣ ∣ \geq 2^{n} \to \infty$ $abs((nr)^n) >= 2^n -> oo$ as $n \to \infty$ $n -> oo$

For what values of rrr does the sequence a_n = (nr)^nan=(nr)na_n = (nr)^n converge ?