# Homework Help: The divergence of a series

1. Jan 20, 2009

### Appa

1. The problem statement, all variables and given/known data

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

2. Relevant equations
The Comparison Test:
0<=ak<=bk
1.The series $$\Sigma$$$$\stackrel{\infty}{k=1}$$ ak converges if the series $$\Sigma$$$$\stackrel{\infty}{k=1}$$ bk converges.
2. The series $$\Sigma$$$$\stackrel{\infty}{k=1}$$ bk diverges if the series $$\Sigma$$$$\stackrel{\infty}{k=1}$$ ak diverges.

3. 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.

2. Jan 20, 2009

### 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

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

3. Jan 20, 2009

### 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?

4. Jan 20, 2009

### 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.

5. Jan 22, 2009

### 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.