Rearranging Series

andrey21

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???

micromass

Yes, this is correct. So what are your values for A and B?
What is

[tex]\frac{1}{(n+1)(n+4)}=...[/tex]?

andrey21

A = 1/3
b= -1/3

micromass

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

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

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

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

micromass

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

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

micromass

Correct!! It seems like you've got it!

andrey21

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

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

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

micromass

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

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

Correct!

andrey21

Great thanks micromass :)