#help for root test for series

11 messages · Page 1 of 1 (latest)

mint topaz
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i dont understand how
|a_n| <= p^(n-N)|a_N|
is obtained. Could someone please enlighten me?

brittle elbowBOT
nimble marsh
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(an/an-1)(an-1/an-2)....product of N terms

gritty wolf
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what you want to consider is that |a_(n + 1)|/|a_n| =< (1 + r)/2 = p

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but there is also something that may lead to the inequality you want

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that is you can pick an N so large than |a_n|/|a_N| =< p

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or this is equivalent to |a_n| =< p|a_N|

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now from the below inequality we get |a_n| =< p^(n - N)|a_N| which seems to be missing

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but you can observe that this also works for |a_(n + 1)|/|a_n|

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i am sure some sort of proof by induction will suffice here