- #1

- 905

- 4

[tex]\Sigma \, ln(\dfrac{(k+1)^{2}}{k(k+2)} )[/tex]

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?

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- Thread starter arunma
- Start date

- #1

- 905

- 4

[tex]\Sigma \, ln(\dfrac{(k+1)^{2}}{k(k+2)} )[/tex]

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?

- #2

- 905

- 4

[tex]\Sigma \, \dfrac{(k+2)}{2^{k}k(k+1)}[/tex]

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

- #3

- 289

- 0

[tex]\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) [/tex]

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

[tex]\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) [/tex]

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 :

[tex]\ln \left( \frac{(k+1)}{k} \right) + \ln \left( \frac{(k+n)}{(k+n+1)} \right)[/tex]

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

- #4

- 905

- 4

Wow, I can't believe I didn't think of that! Thanks a lot for your help.

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