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richyw
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limits. How do I show that \[\lim_{k\to\infty} \frac{(1)^k}{k}=0\]
 one year ago
 one year ago
richyw Group Title
limits. How do I show that \[\lim_{k\to\infty} \frac{(1)^k}{k}=0\]
 one year ago
 one year ago

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shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
Do you know LH rule ?? Where you can differentiate the numerator and denominator.
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
sorry I keep messing up the latex and openstudy is so slow.
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
and yeah I know lh rule could be applied since \(k\in\mathbb{R}\). But how does that help?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
Okay since it is (1) , LH rule won't be applied.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
hmm, so, think like this : (1)^n , where n is any natural number, will always be equal to +1 or 1 right ?
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
agreed. and since \(k\to\infty\) I can see that it's 0. just not sure how to show that in any sort of rigorous way.
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
I need to show it though to show that a series is convergent.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
hmm, Am not too comfortable with that. @phi
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
I thought the definition of a limit L is that for any \( \epsilon >0\), there exists an integer N>0 such that \(  L  x_n < \epsilon \) for all n>N the alternating sign does not affect this definition, because you are only looking at the distance away from L. (0 in this case)
 one year ago
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