Integral Test

**icelated** · April 15th 2012, 03:28 PM

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?

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

then,

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

The solution manual then shows!

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

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!

**Plato** · April 15th 2012, 04:10 PM

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

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

**skeeter** · April 15th 2012, 04:17 PM

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?

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

then,

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

The solution manual then shows!

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

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!

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

chain rule ...

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

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

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

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

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

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

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

you left off the negative sign

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