I have tried using the ratio lemma to tackle this question and also the fact $(n+1)^k \geq 1 + nk$ and I haven't reached an answer. How should I go about solving this problem?
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1Find $\lim\frac{a_{n+1}}{a_n}$ where $a_n=\frac{x^n}{n^k}$ – PNDas Oct 29 '20 at 16:51
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limit is zero when $0<x\leq1$.And limit diverges to infinity when $x>1$. – PNDas Oct 29 '20 at 17:33
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And one more. – rtybase Oct 29 '20 at 17:43
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HINT
$x\left (\frac{ n^{k}}{n+1^{k}} \right )= x\left ( \frac{n}{n+1} \right )^{k}=x\left ( \frac{n+1}{n} \right )^{-k}= x\left ( 1+\frac{1}{n} \right )^{-k}$
領域展開
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no this converge when an+1/an <1 , and you will have to check for an+1/an=1 – 領域展開 Oct 29 '20 at 17:41
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Ah I see so if 0<=x<1 then converges to 0 and if x = 1 then an = 1/n^k so also converges to 0. For x>1 does the sequence just diverge then? – Oct 30 '20 at 11:02
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