# Finding sum of infinite series

Tags:
1. Dec 19, 2016

### Ryaners

[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)

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:

Then when I take the limit, the second term vanishes, leaving just the first term as the final value, i.e.:

Needless to say 48/23 ≠ 3/23
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. Dec 19, 2016

### 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?

3. Dec 19, 2016

### Ryaners

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:

4. Dec 19, 2016

### 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 $16$.

5. Dec 19, 2016

### 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
6. Dec 19, 2016

### Ryaners

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:

Does that even make sense?

7. Dec 19, 2016

### 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).

8. Dec 19, 2016

### 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)}

9. Dec 19, 2016

### Ryaners

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 :)

10. Dec 19, 2016

### Ryaners

Fantastic, thank you! That looks much less difficult than I expected. :)

11. Dec 19, 2016

### 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: $\int_{t = 0}^{10} t^2 + 3t~dt = \left.\frac{t^3}{3} + \frac{3t^2}{2}\right|_0^{10}$

12. Dec 19, 2016

### 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.

13. Dec 19, 2016

### Staff: Mentor

What do you get if you reduce $\frac{3}{4x+3}-\frac{3}{4x+7}$ to a common denominator?

14. Dec 20, 2016

### Ryaners

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 :)

15. Dec 20, 2016

### Ryaners

I get it now, thanks for your input :)