9
$\begingroup$

How do I compute

$$\sum_{r=1}^{\infty} \frac{8r}{4r^4 +1}$$

Calculating first few terms tells me that the sum converges to 2. I have also tried squeezing the term.

$\endgroup$
  • 2
    $\begingroup$ Re: "I have also tried squeezing the term": That's not usually a useful technique with series, because you'd have to squeeze the sequence of partial sums between two sequences that have the same limit. (Squeezing individual terms is meaningless: the sequence of terms has to converge to zero in order for the series to converge at all.) $\endgroup$ – ruakh May 2 '18 at 21:06
23
$\begingroup$

Partial fraction expansion gives us$$\begin{align*} & \sum\limits_{r=1}^n\frac {8r}{4r^4+1}=\sum\limits_{r=1}^n\frac 2{2r^2-2r+1}-\sum\limits_{r=1}^n\frac 2{2r^2+2r+1}\\ & =\left[2+\frac 2{5}+\frac 2{13}+\cdots+\frac 2{2n^2-2n+1}\right]-\left[\frac 2{5}+\frac 2{13}+\cdots+\frac 2{2n^2+2n+1}\right]\end{align*}$$Notice how all but the last fraction in the second sum cancels out with the fractions in the first sum. Continuing on indefinitely until the end gives us$$\sum\limits_{r=1}^n\frac {8r}{4r^4+1}=2-\frac 2{2n^2+2n+1}$$As $n\to\infty$, the fraction tends to zero, so your sum equals$$\sum\limits_{r\geq1}\frac {8r}{4r^4+1}=2$$

EDIT: To find the partial fraction decomposition, we first factor the denominator as a product of two quadratics. This can be done by adding and subtracting $4r^2$ so the quartic factors as$$4r^4+1=(2r^2+2r+1)(2r^2-2r+1)$$Now, the decomposition is set up as$$\frac {8r}{(2r^2+2r+1)(2r^2-2r+1)}=\frac {Ar+B}{2r^2-2r+1}+\frac {Cr+D}{2r^2+2r+1}$$

$\endgroup$
  • 1
    $\begingroup$ I wouldn't have expected partial fractions to work without using Complex numbers: $4r^4+1 = (2r^2+i)(2r^2-i)$ $\endgroup$ – mr_e_man May 3 '18 at 1:26
  • $\begingroup$ @mr_e_man That’s not the only way to factor $4r^4+1$. We can add and subtract $4r^2$ to get$$4r^4+1=(2r^2-2r+1)(2r^2+2r+1)$$ $\endgroup$ – Frank W. May 3 '18 at 1:43
  • $\begingroup$ Yes, I see that it works, I was just surprised. Let's factor it completely (using Complex roots of unity): if $u_8 = \sqrt i = \frac{1+i}{\sqrt2}$ , then $(2r^2+i) = (\sqrt2r-u_8^3)(\sqrt2r-u_8^7)$ , and $(2r^2-i) = (\sqrt2r-u_8^1)(\sqrt2r-u_8^5)$ . Regrouping the factors, $(\sqrt2r-u_8^3)(\sqrt2r-u_8^5) = 2r^2-\sqrt2r(u_8^3+u_8^5)+u_8^8 = 2r^2-\sqrt2r(-\sqrt2)+1 = 2r^2+2r+1$ ; and the other factors, $(\sqrt2r-u_8^7)(\sqrt2r-u_8^1) = 2r^2-\sqrt2r(u_8^7+u_8^1)+u_8^8 = 2r^2-\sqrt2r(\sqrt2)+1 = 2r^2-2r+1$ . $\endgroup$ – mr_e_man May 3 '18 at 2:20
  • $\begingroup$ @mr_e_man Yes that certainly works. I still prefer adding and subtracting $4r^2$ though. $\endgroup$ – Frank W. May 3 '18 at 3:05
7
$\begingroup$

Have you considered partial sum formulas?

Wolframalpha finds:

$\displaystyle \sum_{r=1}^n \dfrac{8r}{4r^4+1} = 2-\dfrac{2}{2n^2+2n+1}$

I'd recommend trying to figure out how it calculated that partial sum. Now, the limit as $n\to \infty$ obviously makes the second term vanish.

$\endgroup$
  • $\begingroup$ I'm sorry but I don't know what partial sum formula is. $\endgroup$ – user138523 May 2 '18 at 14:26
  • $\begingroup$ 1st link on Google doesn't clarify much. I have understood that partial sum is sum of 1st n terms of an AP but what is partial sum formula? $\endgroup$ – user138523 May 2 '18 at 14:28
  • 1
    $\begingroup$ A partial sum formula is: $\displaystyle \sum_{r = 1}^\infty f(r) = \lim_{n \to \infty} \sum_{r=1}^n f(r)$ The right hand side is just taking the first $n$ terms. For example: $\displaystyle \sum_{k=0}^n r^k = \dfrac{1-r^{n+1}}{1-r}$ Taking the limit as $n\to \infty$: $\displaystyle \sum_{k=0}^\infty r^k = \dfrac{1}{1-r}$ $\endgroup$ – InterstellarProbe May 2 '18 at 14:35
4
$\begingroup$

Hint: use that

$$4r^4+1=(2r^2)^2+1=((2r^2)^2+4r^2+1)-4r^2=(2r^2+1)^2-(2r)^2=(2r^2+2r+1)(2r^2-2r+1)$$

Now, using simple fractions,

$\dfrac{8r}{4r^4+1}=\dfrac{8r}{(2r^2+2r+1)(2r^2-2r+1)}=\dfrac{2}{2r^2-2r+1}-\dfrac{2}{2r^2+2r+1}=2\left(\dfrac{1}{2r^2-2r+1}-\dfrac{1}{2r^2+2r+1}\right)$.

The idea is calculate $\sum\limits_{r\in\Bbb N}\frac{1}{2r^2\pm 2r+1}$ using the fact that $$\frac{1}{2r^2\pm 2r+1}=\int^1_0 x^{2r^2\pm 2r}dx$$ and interchanging the sum with the integral.

An example is here: Finding the infinite sum of a rational function using integrals

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.