# Thread: Comparison theorem convergence help.

1. ## 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?

2. 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 $\displaystyle x \ge 1$ ...

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

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

note for $\displaystyle x \ge 1$ ...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. chisigma is correct ... sorry for the error.