# Infinite sum converging to 2

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.

• 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.) – ruakh May 2 '18 at 21:06

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$$\frac {8r}{(2r^2+2r+1)(2r^2-2r+1)}=\frac {Ar+B}{2r^2-2r+1}+\frac {Cr+D}{2r^2+2r+1}$$

• I wouldn't have expected partial fractions to work without using Complex numbers: $4r^4+1 = (2r^2+i)(2r^2-i)$ – mr_e_man May 3 '18 at 1:26
• @mr_e_man That’s not the only way to factor $4r^4+1$. We can add and subtract $4r^2$ to get$$4r^4+1=(2r^2-2r+1)(2r^2+2r+1)$$ – Frank W. May 3 '18 at 1:43
• 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, $(\sqrt2r-u_8^7)(\sqrt2r-u_8^1) = 2r^2-\sqrt2r(u_8^7+u_8^1)+u_8^8 = 2r^2-\sqrt2r(\sqrt2)+1 = 2r^2-2r+1$ . – mr_e_man May 3 '18 at 2:20
• @mr_e_man Yes that certainly works. I still prefer adding and subtracting $4r^2$ though. – Frank W. May 3 '18 at 3:05

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.

• I'm sorry but I don't know what partial sum formula is. – user138523 May 2 '18 at 14:26
• 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? – user138523 May 2 '18 at 14:28
• 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 $n\to \infty$: $\displaystyle \sum_{k=0}^\infty r^k = \dfrac{1}{1-r}$ – InterstellarProbe May 2 '18 at 14:35

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