Friday, 17 November 2017

sequences and series - prove limnrightarrowinftyfracbnnk=infty










I have this sequence with b>1 and k a natural, which diverges:
lim
I need to prove this, with what i have learnt till now from my textbook, my simple step is this:




Since n^2\leq2^n for n>3, i said b^n\geq n^k, so it diverges. Is it right?



I am asking here not just to get the right answer, but to learn more wonderful steps and properties.


Answer



\lim_{n \rightarrow \infty} \frac{b^n}{n^k}=\infty



You can use the root test, too: \lim_{ n\to \infty}\sqrt[\large n]{\frac{b^n}{n^k}} = b>1



Therefore, the limit diverges.







The root test takes the \lim of the n-th root of the term: \lim_{n \to \infty} \sqrt[\large n]{|a_n|} = \alpha.



If \alpha < 1 the sum/limit converges.



If \alpha > 1 the sum/limit diverges.



If \alpha = 1, the root test is inconclusive.



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