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Rearranging Series

by andrey21
Tags: rearranging, series
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micromass
#19
Nov30-10, 11:02 AM
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Yes, this is correct. So what are your values for A and B?
What is

[tex]\frac{1}{(n+1)(n+4)}=...[/tex]?
andrey21
#20
Nov30-10, 11:04 AM
P: 466
A = 1/3
b= -1/3
micromass
#21
Nov30-10, 11:08 AM
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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?
andrey21
#22
Nov30-10, 11:19 AM
P: 466
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:
micromass
#23
Nov30-10, 11:22 AM
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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...
andrey21
#24
Nov30-10, 11:25 AM
P: 466
Well in this case it is the first 3 and last 3 terms which remain correct?
micromass
#25
Nov30-10, 11:30 AM
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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]
andrey21
#26
Nov30-10, 11:31 AM
P: 466
so as k tends to infinty the limit is 11/18.
micromass
#27
Nov30-10, 11:36 AM
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Correct!! It seems like you've got it!
andrey21
#28
Nov30-10, 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??
micromass
#29
Nov30-10, 11:46 AM
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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...
andrey21
#30
Nov30-10, 11:48 AM
P: 466
Yes the sum of 1/n^2 is pi^2/6 correct??
micromass
#31
Nov30-10, 11:51 AM
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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...
andrey21
#32
Nov30-10, 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
micromass
#33
Nov30-10, 11:55 AM
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Correct!
andrey21
#34
Nov30-10, 11:58 AM
P: 466
Great thanks micromass :)
andrey21
#35
Nov30-10, 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?
micromass
#36
Nov30-10, 12:38 PM
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The series is special type of serie. What special kinds of sequences/series have you seen?

Edited because I made a mistake somewhere (:


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