lim n approachs infinity. the intergral from 1 to N (x^-1) dx / the intergral from 1 to N (x^alpha)dx = 0 use the tergral from 1 to n= intergral from 1 to k + intergral form k to n
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Originally Posted by sbacon1991 lim n approachs infinity. the intergral from 1 to N (x^-1) dx / the intergral from 1 to N (x^alpha)dx = 0 use the tergral from 1 to n= intergral from 1 to k + intergral form k to n $\displaystyle \lim_{n\to\infty}\frac{\int_1^n \frac{1}{x}dx}{\int_1^n x^{\alpha}dx}=\lim_{n\to\infty} \frac{\ln{n}}{\frac{n^{\alpha+1}}{\alpha+1}-\frac{1}{\alpha+1}}=...$
using the denition of the Riemann integral can you prove it for?
Originally Posted by sbacon1991 using the denition of the Riemann integral can you prove it for? Introducing a Riemann sum into this problem would complicate it beyond all reason. It calls for a simple application of l'Hopital's rule.
is there any more info which can be given for the question?
Originally Posted by sbacon1991 is there any more info which can be given for the question? Do you know any basic calculus? If not, this is a waste of time going forward.
Last edited by Plato; Nov 27th 2011 at 03:27 PM.
yes i do
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