More fun in 2015, Part 22

Algebra Level 4

$P=\prod_{n=0}^{\infty}\left(1+\frac{1}{2015^{2^n}}\right)$

$P$ can be written as $P=\frac{a}{b}$ for two co-prime positive integers $a$ and $b$ . Enter $a-b$ .

The answer is 1.

1 solution

Otto Bretscher
Oct 31, 2015

We can show by induction on $m$ that $\prod_{n=0}^{m}\left(1+x^{2^n}\right)=\frac{1-x^{2^{m+1}}}{1-x}$ so $\prod_{n=0}^{\infty}\left(1+x^{2^n}\right)=\frac{1}{1-x}$ for $|x|<1$ . For $x=\frac{1}{2015}$ the product comes out to be $P=\frac{a}{b}=\frac{2015}{2014}$ so that $a-b=\boxed{1}$ .

More fun in 2015, Part 22

The answer is 1.

1 solution

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