Comparison theorem convergence help.

• March 5th 2011, 05:35 AM
Kuma
Comparison theorem convergence help.
Use comparison theorem to see if the following converges.

integral from 1 to infinity:

5[((1+(x)^1/2)^1/2)/(x)^1/2]

or in words

5 [root(1+root x)]/root x

This is what I did.
comparison theorem states that for f(x) > g(x) for all x > a, if f(x) converges, so does g(x).

I set f(x) as 5 root (1+x)

is that right?
• March 5th 2011, 06:11 AM
skeeter
Quote:

Originally Posted by Kuma
Use comparison theorem to see if the following converges.

integral from 1 to infinity:

5[((1+(x)^1/2)^1/2)/(x)^1/2]

or in words

5 [root(1+root x)]/root x

This is what I did.
comparison theorem states that for f(x) > g(x) for all x > a, if f(x) converges, so does g(x).

I set f(x) as 5 root (1+x)

is that right?

the constant 5 is inconsequential ... pull it out of the integrand.

note for $x \ge 1$ ...

$\dfrac{\sqrt{1+\sqrt{x}}}{\sqrt{x}} \le \dfrac{\sqrt{\sqrt{x} + \sqrt{x}}}{\sqrt{x}} = \dfrac{\sqrt{2}}{\sqrt[4]{x}}$
• March 5th 2011, 08:03 AM
chisigma
Quote:

Originally Posted by skeeter
the constant 5 is inconsequential ... pull it out of the integrand.

note for $x \ge 1$ ...

$\dfrac{\sqrt{1+\sqrt{x}}}{\sqrt{x}} \le \dfrac{\sqrt{\sqrt{x} + \sqrt{x}}}{\sqrt{x}} = \dfrac{\sqrt{2}}{\sqrt[4]{x}}$

You have demonstrated that the function to be integrated is less than a function the integral of which diverges... so that Your 'demonstration' demonstrates nothing... may be is more useful this inequality, valid for $x>1$...

$\displaystyle \frac{\sqrt{1+\sqrt{x}}}{\sqrt{x}} > \frac{1}{\sqrt{x}}$

Kind regards

$\chi$ $\sigma$
• March 5th 2011, 08:09 AM
skeeter
chisigma is correct ... sorry for the error.