# Let alpha be a real number, alpha > -1 now show that

• Nov 27th 2011, 11:12 AM
sbacon1991
Let alpha be a real number, alpha > -1 now show that
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
• Nov 27th 2011, 11:25 AM
Also sprach Zarathustra
Re: Let alpha be a real number, alpha > -1 now show that
Quote:

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

$\displaystyle \lim_{n\to\infty}\frac{\int_1^n \frac{1}{x}dx}{\int_1^n x^{\alpha}dx}=\lim_{n\to\infty} \frac{\ln{n}}{\frac{n^{\alpha+1}}{\alpha+1}-\frac{1}{\alpha+1}}=...$
• Nov 27th 2011, 01:57 PM
sbacon1991
Re: Let alpha be a real number, alpha > -1 now show that
using the de nition of the Riemann integral can you prove it for?
• Nov 27th 2011, 02:22 PM
Plato
Re: Let alpha be a real number, alpha > -1 now show that
Quote:

Originally Posted by sbacon1991
using the de nition 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.
• Nov 27th 2011, 02:51 PM
sbacon1991
Re: Let alpha be a real number, alpha > -1 now show that
is there any more info which can be given for the question?
• Nov 27th 2011, 02:56 PM
Plato
Re: Let alpha be a real number, alpha > -1 now show that
Quote:

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.
• Nov 27th 2011, 07:14 PM
sbacon1991
Re: Let alpha be a real number, alpha > -1 now show that
yes i do