Thread: Integral Test

I am trying to confirm the integral test can be applied. I just cant see how the solution manual gets there solution.
I think there using the quotient rule but i cant seem to work it out.
Can someone help me work out the problem?

$\displaystyle \sum_{n=1}^\infty \frac 1 {\sqrt {n} ( \sqrt {n} + 1 )} $

then,


Let $\displaystyle f(x) = \frac 1 {\sqrt {x} ( \sqrt {x} + 1 )} $

The solution manual then shows!


$\displaystyle f'(x) = \frac {1+2\sqrt x} { 2x^{3/2} ( \sqrt x + 1 )^2} < 0$

With the denominator squared it does show they use the quotient rule. However, can you please
show me the steps on how they derive the answer?
Because i cant seem to get it. Its kinda a nasty problem.
Thank you so very much!

Originally Posted by icelated

I am trying to confirm the integral test can be applied. I just cant see how the solution manual gets there solution.
I think there using the quotient rule but i cant seem to work it out.
Can someone help me work out the problem?
$\displaystyle \sum_{n=1}^\infty \frac 1 {\sqrt {n} ( \sqrt {n} + 1 )} $

Note that $\displaystyle \int_1^\infty {\frac{{dx}}{{\sqrt x \left( {\sqrt x + 1} \right)}}}$ is the same as $\displaystyle \int_{2}^\infty {\frac{{2du}}{u}} $ if $\displaystyle u=\sqrt{x}+1$.
I have no idea what that solution manual could mean.

Originally Posted by icelated

I am trying to confirm the integral test can be applied. I just cant see how the solution manual gets there solution.
I think there using the quotient rule but i cant seem to work it out.
Can someone help me work out the problem?

$\displaystyle \sum_{n=1}^\infty \frac 1 {\sqrt {n} ( \sqrt {n} + 1 )} $

then,


Let $\displaystyle f(x) = \frac 1 {\sqrt {x} ( \sqrt {x} + 1 )} $

The solution manual then shows!


$\displaystyle f'(x) = \frac {1+2\sqrt x} { 2x^{3/2} ( \sqrt x + 1 )^2} < 0$

With the denominator squared it does show they use the quotient rule. However, can you please
show me the steps on how they derive the answer?
Because i cant seem to get it. Its kinda a nasty problem.
Thank you so very much!

$\displaystyle f(x) = \frac{1}{\sqrt{x}(\sqrt{x}+1)} = \frac{1}{x+\sqrt{x}} = (x + \sqrt{x})^{-1}$

chain rule ...

$\displaystyle f'(x) = -(x+\sqrt{x})^{-2} \cdot \left(1 + \frac{1}{2\sqrt{x}}\right)$

$\displaystyle f'(x) = - \frac{1 + \frac{1}{2\sqrt{x}}}{(x+\sqrt{x})^2}$

$\displaystyle f'(x) = - \frac{1 + \frac{1}{2\sqrt{x}}}{(x+\sqrt{x})^2} \cdot \frac{2\sqrt{x}}{2\sqrt{x}}$

$\displaystyle f'(x) = - \frac{2\sqrt{x} + 1}{2\sqrt{x}(x+\sqrt{x})^2}$

$\displaystyle f'(x) = - \frac{2\sqrt{x} + 1}{2\sqrt{x}[\sqrt{x}(\sqrt{x}+1)]^2}$

$\displaystyle f'(x) = - \frac{2\sqrt{x} + 1}{2\sqrt{x}(\sqrt{x})^2(\sqrt{x}+1)^2}$

$\displaystyle f'(x) = - \frac{2\sqrt{x} + 1}{2x^{3/2}(\sqrt{x}+1)^2} < 0$

you left off the negative sign