# Testing series with ln's for convergence

I'm pretty embarassed that with a math degree I don't know how to do this. But today someone asked me to help him test this series for convergence:

$$\Sigma \, ln(\dfrac{(k+1)^{2}}{k(k+2)} )$$

Maybe I'm just having a brain fart, but I can't think of any series to compare this to, and I obviously can't use the ratio, root, or integral tests. Any suggestions?

## Answers and Replies

One other question, if you guys don't mind. I saw the following series,

$$\Sigma \, \dfrac{(k+2)}{2^{k}k(k+1)}$$

Anyway, it's pretty obvious that this series converges, but is there a way to precisely determine what it converges to?

The first one isn't too hard using the rules of logs when you write them out in a series, realising you're actually considering the total multiplication of all the bits inside the ln( ... ) and then taking the ln.

$$\ln \left( \frac{(k+1)^{2}}{k(k+2)} \right) + \ln \left( \frac{(k+2)^{2}}{(k+1)(k+3)} \right) + \ln \left( \frac{(k+3)^{2}}{(k+2)(k+4)} \right) = \ln \left( \frac{(k+1)(k+3)}{k(k+4)} \right)$$

Notice that the k+2 terms have cancelled along with one factor of the other numerators. Extending this you get

$$\ln \left( \frac{(k+1)^{2}}{k(k+2)} \right) + \ldots + \ln \left( \frac{(k+n)^{2}}{(k+n-1)(k+n+1)} \right) = \ln \left( \frac{(k+1)(k+n)}{k(k+n+1)} \right)$$

Can you see from the 3 term case why it extends to this? All terms cancel except two in the first and two in the last.

Split this up back into two expressions :

$$\ln \left( \frac{(k+1)}{k} \right) + \ln \left( \frac{(k+n)}{(k+n+1)} \right)$$

Take n-> infinity gives the second term as ln(1) = 0. The first term depends on what you're taking k to be. If you start at k=1 then you end up with the total sum as ln(2).

Job's a good 'un :)

I'm not sure about the second one. I'd give it a try, but it's 2.40am and I should be asleep Wow, I can't believe I didn't think of that! Thanks a lot for your help.