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Rearranging Series 
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#19
Nov3010, 11:02 AM

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P: 18,346

Yes, this is correct. So what are your values for A and B?
What is [tex]\frac{1}{(n+1)(n+4)}=...[/tex]? 


#20
Nov3010, 11:04 AM

P: 466

A = 1/3
b= 1/3 


#21
Nov3010, 11:08 AM

Mentor
P: 18,346

Yes, so we have the series
[tex]\sum_{n=1}^{+\infty}{\frac{1}{3(n+1)}\frac{1}{3(4+n)} }[/tex] 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? 


#22
Nov3010, 11:19 AM

P: 466

Alot of the terms cancel out and im left with:
1/3+ 1/6+1/9 1/361/391/42 for first 10 terms: 


#23
Nov3010, 11:22 AM

Mentor
P: 18,346

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... 


#24
Nov3010, 11:25 AM

P: 466

Well in this case it is the first 3 and last 3 terms which remain correct?



#25
Nov3010, 11:30 AM

Mentor
P: 18,346

Indeed. So you've shown that
[tex]\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)}[/tex] Now you can easily take the limit [tex]k\rightarrow +\infty[/tex] 


#26
Nov3010, 11:31 AM

P: 466

so as k tends to infinty the limit is 11/18.



#28
Nov3010, 11:39 AM

P: 466

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?? 


#29
Nov3010, 11:46 AM

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P: 18,346

Do you know what the sum of
[tex]\sum_{n=1}^{+\infty}{\frac{1}{n^2}}[/tex] is?? If you don't know the above sum, then your question is very difficult... 


#30
Nov3010, 11:48 AM

P: 466

Yes the sum of 1/n^2 is pi^2/6 correct??



#31
Nov3010, 11:51 AM

Mentor
P: 18,346

Ah, yes. this is good.
Now, you've got that [tex]\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}}[/tex] You know two of the above series... 


#32
Nov3010, 11:55 AM

P: 466

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 


#34
Nov3010, 11:58 AM

P: 466

Great thanks micromass :)



#35
Nov3010, 12:32 PM

P: 466

I do have another series question micromass:
I have bin given the series: 5/4 + 1 + 4/5 + 16/25+.... It says describe what type of series this is? Shall i try find general formula again? 


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