Question

(Intro Analysis) I'm trying to find the sum of an infinite series by partial fraction decomposition. I'm able to perform the PFD correctly but then don't understand what I need to do to find the value of the series. Problem and working below.

Answer 1

\[\text{Use Partial Fraction Decomposition to find the sum of the series}\]\[\sum_{n=1}^{\infty}\frac{1}{n(n+3)}\]

Answer 2

\[\sum_{n=1}^{\infty}\frac{1}{n(n+3)}=\sum_{n=1}^{\infty}\frac{A}{n}+\frac{B}{(n+3)}=\sum_{n=1}^{\infty}A(n+3)+Bn\]

Answer 3

\[\text{Let n = -3.}\]\[A(-3+3)+B(-3)=1\]\[-3B = 1; \ \ \ B = -\frac{1}{3}.\]
I don't remember the exact reasoning here, but there's some logic like since there are no n terms in the numerator, the term \[(A+B)n\]must be equal to zero. Thus, A = -B, and\[A = \frac{1}{3}\]

Answer 4

From here I can plug these coefficients back in and take their limits, which both go to zero-but how does that help me? This tells me nothing as far as I am aware.

Answer 5

@dan815

Answer 6

@amistre64

Answer 7

your partial is flawed

Answer 8

\[\frac{A}{n}+\frac{B}{m}\ne Am+Bn\]

Answer 9

\[\frac{1}{nm}=\frac{A}{n}+\frac{B}{m}\implies 1=Am+Bn\]

Answer 10

i see that you shortcuted that but made an accounting for it later on ...

Answer 11

let n=0 to solver for A, since B(0)=0

Answer 12

1 = (A+B)n +3A is one form of it if we distribute ... but that doesnt seem useful to me, unless n=0 :)

Answer 13

Yeah, in any case, I think I figured it out after you have the coefficients, just gotta make sure I can get a general term of a new sequence out of it

Answer 14

If you plug into the solved coefficients terms, n=1 and upwards:\[a_1 = 1/4\]\[a_2=1/10\]\[a_3=1/18\]\[a_4=1/28\]\[a_5=1/40\]

Answer 15

I see the pattern, just having trouble putting it into a general term, one min

Answer 16

notice that when we reach n+3, we are subtracting terms that already exist i believe

Answer 17

write it out as the A/n + B/(n+3) structure instead

Answer 18

this is called a telescoping series

Answer 19

I'm not sure what you're talking about, without subtracting, the second fractions I get, in order of n up, are: -1/12, -1/15, /1-18 -1/21, etc

Answer 20

\[B = -A\]right?

Answer 21

Yeah

Answer 22

+A/1 - A/(1+3)
+A/2 -A/(2+3)
+A/3 -A/(3+3)
+A/4 -A/(4+3)

adding this up we get
A/4 -A/(1+3) = 0

in fact every n greater than 3 has a negation to it .

Answer 23

1 - 4
2 - 5
3 - 6
4 - 7
5 - 8


1
2
3
4 - 4
5 - 5
6 - 6
7 - 7
8 - 8
...

Answer 24

so the sum is just the first 3 terms, since they do not recieve a negating term

Answer 25

spose set A=(1,2,3,4,5,6,...)

set B=(4,5,6,...)

A-B = (1,2,3)

right?

Answer 26

Yeah, I see it, just had to write it out and perform the cancellations. The only remaining terms are A/1, A/2. and A/3.

Answer 27

then thats the sum :)

Answer 28

How can I otherwise prove that or show that? While that makes sense it doesn't feel very rigorous

Answer 29

proof it by a telescopic series.

Answer 30

Like the way with the 1/4, 1/10, 1/18, etc, it's 1/4, 1/4+6, 1/4+6+8, etc

What?

Answer 31

http://tutorial.math.lamar.edu/Classes/CalcII/Series_Special.aspx

Answer 32

an inductive proof might be useful

Answer 33

Yeah, since this is analysis I feel like he won't let me go with, "they cancel out, so this is the sum", I'm not totally sure how he wants me to do this the way he wants

Answer 34

show that for every term n>3, there is a cancelation. or are you wanting to develop a partial sum series?

Answer 35

im not keen on methods tho

Answer 36

http://www.math.wisc.edu/~park/Fall2011/sequences/telescoping.pdf

maybe?

Answer 37

hi sry im here, r u done