Use partial fractions to find the sum of the series

In summary, to use partial fractions to find the sum of a series, the denominator of the series expression must be factored and a system of equations using the coefficients of each term in the partial fraction decomposition must be set up. The purpose of using partial fractions is to simplify complex series expressions and make them easier to integrate and evaluate. However, partial fractions can only be used for series with rational functions in the denominator, and the series must have a finite number of terms. Additionally, it may not always be possible to find the exact sum of a series using partial fractions and approximations may need to be used in some cases.
  • #1
JFUNR
5
0

Homework Statement


Use partial fractions to find the sum of the series: [itex]\Sigma[/itex]n=1 to infinity [itex]\frac{5}{n(n+1)(n+2}[/itex]


Homework Equations


Partial Fraction breakdown: [itex]\Sigma[/itex] [itex]\frac{5}{2n}[/itex]+[itex]\frac{5}{2(n+2)}[/itex]+[itex]\frac{5}{(n+1)}[/itex]


The Attempt at a Solution


When I tried to cancel terms out, it is erratic and no pattern seems to emerge. It is not geometric so using a/(1-r) isn't possible. When I separated the terms, each diverges based on the nth term test. According to the all-knowing wolfram-alpha, the series converges to 5/4...I feel as if I am missing something big in my attempts at solving this one? any help?
 
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  • #2
JFUNR said:

Homework Statement


Use partial fractions to find the sum of the series: [itex]\Sigma[/itex]n=1 to infinity [itex]\frac{5}{n(n+1)(n+2}[/itex]


Homework Equations


Partial Fraction breakdown: [itex]\Sigma[/itex] [itex]\frac{5}{2n}[/itex]+[itex]\frac{5}{2(n+2)}[/itex]+[itex]\frac{5}{(n+1)}[/itex]


The Attempt at a Solution


When I tried to cancel terms out, it is erratic and no pattern seems to emerge. It is not geometric so using a/(1-r) isn't possible. When I separated the terms, each diverges based on the nth term test. According to the all-knowing wolfram-alpha, the series converges to 5/4...I feel as if I am missing something big in my attempts at solving this one? any help?

Your partial fraction expansion is incorrect.

RGV
 
  • #3
Ray Vickson said:
Your partial fraction expansion is incorrect.

RGV

ah that was a typo, the partial fraction expansion I have in my attempts is...[itex]\frac{5}{2n}[/itex]+[itex]\frac{5}{2(n+2)}[/itex]-[itex]\frac{5}{n+1}[/itex]

I still have the same issue..
 
  • #4
Edit : Whoops you had the right expansion I didn't see your post there.

There will be a pattern you will notice, but a more important question to ask yourself is : Suppose your sum is going from 1 to infinity. Let's say you're summing some A+B+C which are all composed of a variable n ( the variable you are summing of course ). If one of A, B, or C diverges, does the whole series diverge?
 
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  • #5
Hint 1: [tex] \sum_{n=1}^{\infty} f(n) = \lim_{N \to \infty} \sum_{n=1}^{N} f(n)[/tex]

Hint 2: [tex]\sum_{n=1}^{N} \frac{1}{n+2} = \left( \sum_{n=1}^{N+2} \frac{1}{n} \right) - \frac{1}{1} - \frac{1}{2}[/tex]
 
  • #6
reindex second and third sums.
and notice you get a whole lot of cancelation.
 
  • #7
qbert said:
reindex second and third sums.
and notice you get a whole lot of cancelation.

How do you mean?
 
  • #8
JFUNR said:
How do you mean?

He means that you should write it out in detail and see what happens.

RGV
 
  • #9
Ray Vickson said:
He means that you should write it out in detail and see what happens.

RGV

by all means...enlighten me..where's the pattern...

([itex]\frac{5}{2}[/itex]+[itex]\frac{5}{6}[/itex]-[itex]\frac{5}{2}[/itex])+([itex]\frac{5}{4}[/itex]+[itex]\frac{5}{8}[/itex]-[itex]\frac{5}{3}[/itex])+([itex]\frac{5}{6}[/itex]+[itex]\frac{5}{10}[/itex]-[itex]\frac{5}{4}[/itex])+([itex]\frac{5}{8}[/itex]+[itex]\frac{5}{12}[/itex]-[itex]\frac{5}{5}[/itex])+([itex]\frac{5}{10}[/itex]+[itex]\frac{5}{14}[/itex]-[itex]\frac{5}{6}[/itex])+([itex]\frac{5}{12}[/itex]+[itex]\frac{5}{16}[/itex]-[itex]\frac{5}{7}[/itex])+([itex]\frac{5}{14}[/itex]+[itex]\frac{5}{18}[/itex]-[itex]\frac{5}{8}[/itex])+...+

while some terms do cancel, I do not see a patter out of the first 7 terms...but like I said...enlighten me
 
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  • #10
JFUNR said:
by all means...enlighten me..where's the pattern...

([itex]\frac{5}{2}[/itex]+[itex]\frac{5}{6}[/itex]-[itex]\frac{5}{2}[/itex])+([itex]\frac{5}{4}[/itex]+[itex]\frac{5}{8}[/itex]-[itex]\frac{5}{3}[/itex])+([itex]\frac{5}{6}[/itex]+[itex]\frac{5}{10}[/itex]-[itex]\frac{5}{4}[/itex])+([itex]\frac{5}{8}[/itex]+[itex]\frac{5}{12}[/itex]-[itex]\frac{5}{5}[/itex])+([itex]\frac{5}{10}[/itex]+[itex]\frac{5}{14}[/itex]-[itex]\frac{5}{6}[/itex])+([itex]\frac{5}{12}[/itex]+[itex]\frac{5}{16}[/itex]-[itex]\frac{5}{7}[/itex])+([itex]\frac{5}{14}[/itex]+[itex]\frac{5}{18}[/itex]-[itex]\frac{5}{8}[/itex])+...+

while some terms do cancel, I do not see a patter out of the first 7 terms...but like I said...enlighten me

Do you really need someone to tell you that
[tex] \sum_{k=1}^N \frac{1}{2k} = \frac{1}{2} \sum_{k=1}^N \frac{1}{k}\;?[/tex]

RGV
 
  • #11
JFUNR said:
by all means...enlighten me..where's the pattern...

([itex]\frac{5}{2}[/itex]+[itex]\frac{5}{6}[/itex]-[itex]\frac{5}{2}[/itex])+([itex]\frac{5}{4}[/itex]+[itex]\frac{5}{8}[/itex]-[itex]\frac{5}{3}[/itex])+([itex]\frac{5}{6}[/itex]+[itex]\frac{5}{10}[/itex]-[itex]\frac{5}{4}[/itex])+([itex]\frac{5}{8}[/itex]+[itex]\frac{5}{12}[/itex]-[itex]\frac{5}{5}[/itex])+([itex]\frac{5}{10}[/itex]+[itex]\frac{5}{14}[/itex]-[itex]\frac{5}{6}[/itex])+([itex]\frac{5}{12}[/itex]+[itex]\frac{5}{16}[/itex]-[itex]\frac{5}{7}[/itex])+([itex]\frac{5}{14}[/itex]+[itex]\frac{5}{18}[/itex]-[itex]\frac{5}{8}[/itex])+...+

while some terms do cancel, I do not see a patter out of the first 7 terms...but like I said...enlighten me

The pattern may indeed be a little difficult to see like this, but notice, for example, that [itex]\frac{5}{6}+\frac{5}{6}-\frac{5}{3}=0[/itex] and [itex]\frac{5}{8}+\frac{5}{8}-\frac{5}{4}=0[/itex].

To formally reorganize the sum so that the cancellations become obvious, first look at the finite sum and apply the second hint I gave you to both the 5/(2(n+2)) and -5(n+1) terms. After reorganizing the finite sum to get the desired cancellations, take the limit as N->infinity.
 

Related to Use partial fractions to find the sum of the series

1. How do you use partial fractions to find the sum of a series?

To use partial fractions to find the sum of a series, you first need to factor the denominator of the series expression. Then, you set up a system of equations using the coefficients of each term in the partial fraction decomposition. Solving this system of equations will give you the values of the unknown constants, which can then be substituted back into the partial fraction decomposition. The resulting expression can be integrated and evaluated to find the sum of the series.

2. What is the purpose of using partial fractions in finding the sum of a series?

The purpose of using partial fractions is to simplify complex series expressions into more manageable terms that can be easily integrated and evaluated. This technique is particularly useful for series with rational functions as it allows us to break them down into simpler, known functions.

3. Can you use partial fractions to find the sum of any series?

No, partial fractions can only be used to find the sum of series with rational functions in the denominator. If the series has a different form, such as trigonometric functions or exponential functions, then other techniques may need to be used to find the sum.

4. How do you know if a series can be solved using partial fractions?

A series can be solved using partial fractions if it can be expressed as a rational function, where the denominator can be factored into linear or quadratic terms. If the series cannot be written in this form, then partial fractions cannot be used to find the sum.

5. Are there any limitations to using partial fractions to find the sum of a series?

Yes, there are some limitations to using partial fractions. It can only be used for series with rational functions in the denominator, and the series must have a finite number of terms. Additionally, it may not always be possible to find the exact sum of a series using partial fractions, and in these cases, approximations may need to be used.

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