1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding sum of infinite series

  1. Dec 19, 2016 #1
    [Please excuse the screengrabs of the fomulae - I'll get around to learning TeX someday!]

    1. The problem statement, all variables and given/known data

    Find the sum of this series (answer included - not the one I'm getting)
    3lz8gX9.png

    3. The attempt at a solution
    So I'm trying to sum this series as a telescoping sum. I decomposed the fraction and the partial sum collapses down to this:
    PkZ9pxR.png
    Then when I take the limit, the second term vanishes, leaving just the first term as the final value, i.e.:
    t3SRR8e.png
    Needless to say 48/23 ≠ 3/23 :redface:
    I can't see what I'm doing wrong - I've triple-checked everything & must be misunderstanding something fundamentally speaking. The partial fraction decomposition is (to the best of my / Wolfram A's knowledge) correct. Can anyone point out my mistake(s)?
     
  2. jcsd
  3. Dec 19, 2016 #2

    fresh_42

    Staff: Mentor

    It's hidden in the part you didn't show us. Can you explain your decomposition of ##16x^2+40x+21## into two fractions?
     
  4. Dec 19, 2016 #3
    Sure thing. I used the quadratic formula to find the roots of the polynomial in the denominator, & then decomposed the fraction - here's the process in detail:
    tQhfcLj.png
     
  5. Dec 19, 2016 #4

    fresh_42

    Staff: Mentor

    Yes, ##16x^2+40x+21=(4x+7)(4x+3)## and this gives the new denominators. But from the first to the second line is an unseen error, because the first equality in the second line is wrong by a factor ##16##.
     
  6. Dec 19, 2016 #5

    Mark44

    Staff: Mentor

    You have a mistake right off the bat.
    $$\frac{12}{16x^2 + 40x + 21} \ne \frac{12}{(x + 3/4)(x + 7/4)}$$
    You have lost two factors of 4 from the denominators on the right.
     
    Last edited: Dec 19, 2016
  7. Dec 19, 2016 #6
    Ok, I can see that now - so was my mistake to use the quadratic formula the same way I'd use it if the polynomial = 0, where I was kind of unconsciously dividing through by the 4 without realising? In other words, I should have:

    QWCYjjH.png

    Does that even make sense? :bugeye:
     
  8. Dec 19, 2016 #7

    fresh_42

    Staff: Mentor

    No, this part of your calculation has been correct. You can work either with ##x+\frac{3}{4}\, , \,x+\frac{7}{4}## or with ##4x+3\, , \,4x+7##.
    However, this has an effect to your nominators. Look again at the inequality @Mark44 pointed to in post #5 - make it an equality (with either denominators).
     
  9. Dec 19, 2016 #8

    Mark44

    Staff: Mentor

    We have a tutorial here: https://www.physicsforums.com/help/latexhelp/
    You can be up an running in a short time. To see what I or fresh_42 did, right-click on any of the expressions we wrote, and choose "Show Math As.. " and "TeX Commands". This will show the LaTeX script we wrote.

    In post #5 I wrote this:
    ##\frac{12}{16x^2 + 40x + 21} \ne \frac{12}{(x + 3/4)(x + 7/4)}##

    The TeX script looks like this:
    \frac{12}{16x^2 + 40x + 21} \ne \frac{12}{(x + 3/4)(x + 7/4)}
     
  10. Dec 19, 2016 #9
    Right - so I should recognise once I find the roots that I might need to scale them by a third (constant) factor so I'm not changing the value of the expression? Just trying to get to the bottom of what mental shortcut I'm taking that I shouldn't be.

    I had assumed that because the quadratic formula incorporated all the coefficients 'as they are' (i.e. a=16, b=40, c=21 in this case), the roots it 'spits out' would have included all that info. 'Course when I look at it I can see that the two expressions aren't equal (as pointed out by Mark44). I'm unclear on why that's the case, though. How can I make sure when I'm performing this kind of operation in future that I haven't magically eliminated a constant factor, like I did here?

    Thanks for your help so far by the way :)
     
  11. Dec 19, 2016 #10
    Fantastic, thank you! That looks much less difficult than I expected. :)
     
  12. Dec 19, 2016 #11

    Mark44

    Staff: Mentor

    Fractions aren't too complicated. Here are a couple of others that I frequently use:

    Summations:
    \sum_{n = 1}^{\infty} \frac 1 {n^2}
    In rendered form: ##\sum_{n = 1}^{\infty} \frac 1 {n^2}##

    Integrals:
    \int_{t = 0}^{10} t^2 + 3t~dt = \left.\frac{t^3}{3} + \frac{3t^2}{2}\right|_0^{10}
    In rendered form: ##\int_{t = 0}^{10} t^2 + 3t~dt = \left.\frac{t^3}{3} + \frac{3t^2}{2}\right|_0^{10}##
     
  13. Dec 19, 2016 #12

    fresh_42

    Staff: Mentor

    This is a matter of taste. Personally I like to work with equations ##1\cdot x^2 + px +q = 0## because I don't have to bother the scaling then.
    I also use ##x_{1,2} = -\frac{p}{2} \pm \sqrt{(-\frac{p}{2})^2-q}## because I learnt it this way, although mostly ##x_{1,2} = -\frac{1}{2}(p \pm \sqrt{p^2-4q})## is used. Once you have the decomposition, say ##(x+\frac{3}{4})\cdot (x+\frac{7}{4})##, simply make sure that you decomposed ##16x^2+40x+21=16(x^2+\frac{5}{2}+\frac{21}{16})= 16 \cdot (x+\frac{3}{4})\cdot (x+\frac{7}{4})## and not ##16x^2+40x+21=(4x+3)\cdot (4x+7)##. Until then and after this everything has been o.k. with your calculation. Maybe it's easiest to take the factor ##\frac{12}{16}## completely in front of all the rest and multiply the result at last again with it. This way the algorithmic procedure doesn't have to consider the scaling factor. But as I said above: it's a matter of taste. Find a way that minimizes potential mistakes for you.
     
  14. Dec 19, 2016 #13
    What do you get if you reduce ##\frac{3}{4x+3}-\frac{3}{4x+7}## to a common denominator?
     
  15. Dec 20, 2016 #14
    Yes! Ok! This is making sense now. So the safest thing to do is to put the quadratic in a form where the coefficient of the x2 term is 1 before applying the quadratic formula, so you don't lose track of any constant factors while finding the roots. That's exactly the insight I was looking for, thanks for your help :)
     
  16. Dec 20, 2016 #15
    I get it now, thanks for your input :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding sum of infinite series
  1. Infinite sums/series (Replies: 5)

  2. Sum of a infinite series (Replies: 13)

Loading...