Partial Fractions Sum of Series

[whatlifeforme]

Homework Statement

Use partial fractions to find the sum of the series.

Homework Equations

\displaystyle \sum^{∞}_{n=1} \frac{8}{n(n+3)}

The Attempt at a Solution

I end up with:

\displaystyle \frac{8}{3n} - \frac{8}{3(n+3)}

I am stuck here.

 

[ArcanaNoir]

Did you try writing out some terms of the series? I took it to n=7 to get a good feeling.

 

[tiny-tim]

hi whatlifeforme!

ok, now try writing out the first few terms, and see what you notice

(if that doesn't help, try it with n(n+1) instead of n(n+3), and then adapt)

 

[whatlifeforme]

i'm still confused. after 7 terms, i get to 3.99 but it looks like it is still increasing.

 

[ArcanaNoir]

Ack, decimals are bad :) Use actual fractions, and write it out like this:
\frac{8}{3}-\frac{8}{12}+\frac{8}{6}-\frac{8}{15}+\dots
Continue this pattern to at least n=7. You might want to look up "telescoping series".

 

[whatlifeforme]

i have yet to master telescoping series. any help please?

I have gone to n=7 and have 6 terms left that have not be canceled.

 

[tiny-tim]

and what are the 6 terms?

 

[whatlifeforme]

8/3 + 8/6 + 8/9 - 8/24 - 8/27 - 8/30

 

[Ray Vickson]

Homework Statement

Use partial fractions to find the sum of the series.

Homework Equations

\displaystyle \sum^{∞}_{n=1} \frac{8}{n(n+3)}

The Attempt at a Solution

I end up with:

\displaystyle \frac{8}{3n} - \frac{8}{3(n+3)}

I am stuck here.

Evaluate the finite sum
S_N = \sum_{n=1}^N \frac{8}{n(n+3)}, then take the limit as N → ∞. Your partial fraction representation makes this straightforward.

 

[SammyS]

8/3 + 8/6 + 8/9 - 8/24 - 8/27 - 8/30

Write them out without cancelling them.

 

[whatlifeforme]

(8/3 - 8/13) + (8/6 - 8/15) + (8/9 -8/15) + (8/12 - 8/21) + (8/15 - 8/24) + (8/18 - 8/27) + (8/21 - 8/30)

 

[ArcanaNoir]

Here is an example. Suppose my series came out like this:
\frac{1}{2} - \frac{1}{5} +\frac{1}{3} - \frac{1}{6} + \frac{1}{4} - \frac{1}{7} + \frac{1}{5} - \frac{1}{8} + \frac{1}{6} + \dots

Notice that this can be rearranged as \frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\frac{1}{5} - \frac{1}{5} +\frac{1}{6} - \frac{1}{6} \dots
See that every fraction has a positive and a negative, so everything cancels except for the first few fractions, 1/2, 1/3, and 1/4. The sum of this series is \frac{1}{2}+\frac{1}{3}+\frac{1}{4}.

 

[ArcanaNoir]

(8/3 - 8/13) + (8/6 - 8/15) + (8/9 -8/15) + (8/12 - 8/21) + (8/15 - 8/24) + (8/18 - 8/27) + (8/21 - 8/30)

Can you form pairs of positive and negative like -8/15 + 8/15 ?

 

[SammyS]

(8/3 - 8/12) + (8/6 - 8/15) + (8/9 -8/15) + (8/12 - 8/21) + (8/15 - 8/24) + (8/18 - 8/27) + (8/21 - 8/30)

How about like this ?

8/3 + 8/6 + 8/9 + 8/12 + 8/15 + 8/18 + 8/21 + ...

- 8/12 - 8/15 - 8/18 - 8/21 - 8/24 - 8/27 - 8/30 - ...

Do you see what happens?

 

[whatlifeforme]

so I'm left with 8/3 + 8/6 + 8/9

 

[ArcanaNoir]

Exactly :) Does it make sense?

 

[whatlifeforme]

not yet. i still don't have the sum. doesn't it need to look like sum = 8/3 - 1/k-1

then taking the limit it would be 8/3.

this isn't valid for this problem just an example.

i know when writing out 1/n - 1/n-1 type stuff.. i always had a problem with telescoping series.

 

[ArcanaNoir]

The sum is a number. Specifically the sum is the sum of 8/3+8/6+8/9.

 

[whatlifeforme]

so the answer is 44/9.

 

[ArcanaNoir]

Right.

 

[whatlifeforme]

check out this video on telescoping series. he does not include all the left over terms but converts them into 1/k+1 - 1/k+2

 

[ArcanaNoir]

Yes, but then he takes the limit as k goes to infinity and both of those terms become 0.

 

[SammyS]

check out this video on telescoping series. he does not include all the left over terms but converts them into 1/k+1 - 1/k+2

OK.

Writing your series like he does $\displaystyle \ \lim_{k\to\infty} \sum_{n=1}^{k} \frac{8}{n(n+3)} \ .$

What are those last three "left over" terms?