It's actually two fractions

Algebra Level 4

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

If the series above equals to $\dfrac ab$ for coprime positive integers $a$ and $b$ , find the value of $a + b$ .

The answer is 11.

2 solutions

Noel Lo
Jun 18, 2015

$\frac{r}{r^4+4} = \frac{r}{r^4+4r^2+4 - 4r^2} = \frac{r}{(r^2+2)^2 - (2r)^2}$

$= \frac{r}{(r^2 - 2r+2)(r^2 +2r+2)}= \frac{1}{4}(\frac{1}{r^2 - 2r+2} - \frac{1}{r^2 + 2r + 2})$

When we apply telescoping sums we realise that $(r-2)^2 + 2(r-2) + 2 = r^2 - 2r + 2$ so we can cancel out recurring terms and get $\frac{1}{4}(\frac{1}{1-2+2} + \frac{1}{4-4+2}) = \frac{1}{4}(1+\frac{1}{2}) = \frac{1}{4} (\frac{3}{2}) = \frac{3}{8}$

So we have $a+b =3+8 = \boxed{11}$

Moderator note:

Yes, telescoping sum works here. Note that $r^4 + 4$ is in the form of $a^4 + 4b^4$ , thus it can be factorize by Sophie Germain Identity.

Nice work, Noel!

Niceeeee solution, very niceeee

Muhammad Ardivan - 5 years, 12 months ago

Realising that the denominator can be facorised easily by Sophie Germain's identity makes telescoping sums the hero once again.

Devin Ky - 5 years, 11 months ago

In second line , how have you simplified $=\frac{r}{(r^2-2r+2)(r^2+2r+2)}$ to $\frac{1}{4}$ $(\frac{1}{r^2-2r+2}-\frac{1}{r^2+2r+2})$ ?? Please explain with steps . How you got inspired ??

Chirayu Bhardwaj - 5 years, 2 months ago

Very nice!

Simon Boutin - 5 years, 12 months ago

Ramiel To-ong
Jun 19, 2015

telescoping sums really works.

It's actually two fractions

The answer is 11.

2 solutions

Moderator note:

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