The Convergence of the Series (sqrt(k+1) - sqrt(k))/k

[Appa]

Homework Statement

Is the series \Sigma\stackrel{\infty}{k=1} (\sqrt{k+1} - \sqrt{k})/k convergent or divergent?

Homework Equations

The Comparison Test:
0<=a_k<=b_k
1.The series \Sigma\stackrel{\infty}{k=1} a_k converges if the series \Sigma\stackrel{\infty}{k=1} b_k converges.
2. The series \Sigma\stackrel{\infty}{k=1} b_k diverges if the series \Sigma\stackrel{\infty}{k=1} a_k diverges.

The Attempt at a Solution

I computed the equation until it looked like this: \Sigma\stackrel{\infty}{k=1} (\sqrt{1/k + 1/k^2} - 1/\sqrt{k}) and then I tried to find some other series that would be smaller than the original but still diverge because my guess is that this series diverges. But if I take the series \Sigma\stackrel{\infty}{k=1} (\sqrt{1/k^2} - 1/\sqrt{k}) the terms of the series become negative and the rules of the Comparison Test don't apply.
Then I tried the series \Sigma\stackrel{\infty}{k=1} (\sqrt{k+1} - \sqrt{k-1})/k and computed it to this: \Sigma\stackrel{\infty}{k=1} (\sqrt{1/k + 1/k^2} - \sqrt{1/k-1/k^2}) but it's even harder to analyse than the original series.
Any hints? I also tried series that are greater than the original but found all of them divergent so they weren't of any help.

 

[NoMoreExams]

Personally I would do integral test. Your terms are decreasing, I believe; you know that

\lim_{k=\infty} f(k) = 0

So you'd have to evaluate

\int_{1}^{\infty} \frac{\sqrt{k+1} - \sqrt{k}}{k} dk

The first part of that integral is tricky and I don't see a quick way to do it but I believe the integral evaluates to


\int_{1}^{\infty} \frac{\sqrt{k+1} - \sqrt{k}}{k} dk = 2 - 2\sqrt{2} + ln\left|\frac{\sqrt{2}+1}{\sqrt{2}-1}\right|

 

[NoMoreExams]

The best I can come up with is that

\frac{\sqrt{k+1}}{k} dk = \frac{k+1}{k\sqrt{k+1}} dk

So if we let

u = \sqrt{k+1} \Rightarrow 2du = \frac{dk}{\sqrt{k+1}}

And we are left with integrating

2\frac{u^2}{u^{2} - 1} du

Which you can simplify fairly easily using polynomial division and then do partial fractions on what's left.

Does that help?

 

[Dick]

You can do a comparison test as well. First multiply numerator and denominator by sqrt(k+1)+sqrt(k). Now compare it with a p-series.

 

[Appa]

Now I was able to compute the series \Sigma\stackrel{\infty}{k=1} (\sqrt{k+1} - \sqrt{k})/k to \Sigma\stackrel{\infty}{k=1} 1/k(\sqrt{k+1} + \sqrt{k}).

And from there I was able to tell that 1/k(\sqrt{k+1} + \sqrt{k}) < 1/k(2\sqrt{k}) . And because \Sigma\stackrel{\infty}{k=1} 1/k(2\sqrt{k}) converges, the Comparison Test states that \Sigma\stackrel{\infty}{k=1} (\sqrt{k+1} - \sqrt{k})/k must converge too.

Thank-you for your help!