## Rearranging Series

Ah ok so that gives n^2 +5n +4 which factors to:

(n+1)(n+4)

Then solving will give values for A and B. Then wot shall i do???
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Yes, this is correct. So what are your values for A and B? What is $$\frac{1}{(n+1)(n+4)}=...$$?
 A = 1/3 b= -1/3
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Yes, so we have the series $$\sum_{n=1}^{+\infty}{\frac{1}{3(n+1)}-\frac{1}{3(4+n)} }$$ Now you have to use a little trick. Try to write out this series for the first 10 terms (without adding any of the fractions). Do you see something cool?
 Alot of the terms cancel out and im left with: 1/3+ 1/6+1/9 -1/36-1/39-1/42 for first 10 terms:
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Yes, that's exactly what we wanted! Can you do this in general now? If you calculate the first k terms, there will be a lot of terms cancelling out. So which of the terms remain? If you don't see this immediately, try taking k some other values. Try taking k=20 and k=30. Then you will be ready to handle the general situation...
 Well in this case it is the first 3 and last 3 terms which remain correct?
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Indeed. So you've shown that $$\sum_{n=1}^{k}{\left(\frac{1}{3(n+1)}-\frac{1}{3(n+4)}\right)}=\frac{1}{3}+\frac{1}{6}+\frac{1}{9}-\frac{1}{3(k+2)}-\frac{1}{3(k+3)}-\frac{1}{3(k+4)}$$ Now you can easily take the limit $$k\rightarrow +\infty$$
 so as k tends to infinty the limit is 11/18.
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Correct!! It seems like you've got it!
 Fantastic thanks micromass. I do have one final question very similar. given the sequence: 1+1/9+1/25+1/49.... I know this can be written as: sum 1/(2n+1)^2 correct?? What does this converge to??
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Do you know what the sum of $$\sum_{n=1}^{+\infty}{\frac{1}{n^2}}$$ is?? If you don't know the above sum, then your question is very difficult...
 Yes the sum of 1/n^2 is pi^2/6 correct??
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Ah, yes. this is good. Now, you've got that $$\sum_{n=1}^{+\infty}{\frac{1}{n^2}}=\sum_{n=1}^{+\infty}{\frac{1}{(2n)^ 2}}+\sum_{n=1}^{+\infty}{\frac{1}{(2n+1)^2}}$$ You know two of the above series...
 ah yes i see simply solve to get: 1/4n^2 which equals pi^2/24 correct?? Then subtract to leave me value for 1/(2n+1)^2
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Correct!
 Great thanks micromass :)