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

**sbacon1991** · November 27th 2011, 11:12 AM

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

**Also sprach Zarathustra** · November 27th 2011, 11:25 AM

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

$\lim_{n\to\infty}\frac{\int_1^n \frac{1}{x}dx}{\int_1^n x^{\alpha}dx}=\lim_{n\to\infty} \frac{\ln{n}}{\frac{n^{\alpha+1}}{\alpha+1}-\frac{1}{\alpha+1}}=...$

**sbacon1991** · November 27th 2011, 01:57 PM

using the denition of the Riemann integral can you prove it for?

**Plato** · November 27th 2011, 02:22 PM

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.

**sbacon1991** · November 27th 2011, 02:51 PM

is there any more info which can be given for the question?

**Plato** · November 27th 2011, 02:56 PM

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.

**sbacon1991** · November 27th 2011, 07:14 PM

yes i do

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

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

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