A product of our times

Calculus Level 3

$\large \displaystyle\prod_{n=0}^{\infty} \left( 1 + \dfrac{1}{2015^{2^{n}}} \right) = \dfrac{a}{b}$

For the equation above, given that $a$ and $b$ are positive coprime integers, find $a + b.$

The answer is 4029.

1 solution

Brian Charlesworth
May 5, 2015

Multiply the product on the left-hand side of the equation by $1$ in the form

$\dfrac{1 - \dfrac{1}{2015}}{1 - \dfrac{1}{2015}}.$ The product will then be

$\dfrac{2015}{2014} \left(1 - \dfrac{1}{2015^{2^{1}}}\right)\left(1 + \dfrac{1}{2015^{2^{1}}}\right)\left(1 + \dfrac{1}{2015^{2^{2}}}\right)....... =$

$\dfrac{2015}{2014} \left(1 - \dfrac{1}{2015^{2^{2}}}\right)\left(1 + \dfrac{1}{2015^{2^{2}}}\right)\left(1 + \dfrac{1}{2015^{2^{3}}}\right)....$

As can be seen, this product "compresses" one pair at a time, since for any integer $n \ge 0$ we have that

$\left(1 - \dfrac{1}{2015^{2^{n}}}\right)\left(1 + \dfrac{1}{2015^{2^{n}}}\right) = 1 - \dfrac{1}{2015^{2^{n+1}}}.$

This "compressing" continues on ad infinitum, and thus the product goes to

$\dfrac{2015}{2014} \lim_{n \rightarrow \infty} \left(1 \pm \dfrac{1}{2015^{2^{n}}}\right) = \dfrac{2015}{2014}.$

Thus $a + b = 2015 + 2014 = \boxed{4029}.$

Moderator note:

Splendid. Is the generalization $\displaystyle \prod_{n=0}^\infty \left(1+\frac1{A^{2^n}}\right)=1+\frac1{A-1}$ true for all $A>1$ ?

If we're considering this:

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

Then, I think that the following limit is more justified:

$\large\mathcal{P}=\left(1-\frac{1}{2015}\right)^{-1}\cdot\lim_{n\to\infty}\left(1-\frac{1}{2015^{2^{n+1}}}\right)=\frac{2015}{2014}$

Although this doesn't change much.

Prasun Biswas - 6 years, 1 month ago

You're right. I got fast and loose with the limit at the end since I wasn't quite sure how to present it, but the answer is still the same regardless.

Brian Charlesworth - 6 years, 1 month ago

This is off topic. Is $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin ^{ 2 }{ (nx) } }{ \sin ^{ 2 }{ (x) } } dx } =\frac { n\pi }{ 2 }$ true ?

Vighnesh Raut - 6 years, 1 month ago

@Vighnesh Raut – This beautiful identity does appear to be true. This link shows that

$\displaystyle\int_{0}^{2\pi} \dfrac{\sin^{2}(nx)}{\sin^{2}(x)} dx = 2n\pi.$

From here we can use a symmetry argument to conclude that the integral from $0$ to $\dfrac{\pi}{2}$ is $\dfrac{1}{4}*2n\pi = \dfrac{n\pi}{2}.$

Brian Charlesworth - 6 years, 1 month ago

@Brian Charlesworth – So, is it possible to prove it by induction because when I tried , it went long up to 2 pages and I reached no where.

Vighnesh Raut - 6 years, 1 month ago

@Vighnesh Raut – There are already three answers posted in that Math SE link he gave. You should check that out. At times, a direct proof is easier and more doable than an inductive proof.

Prasun Biswas - 6 years, 1 month ago

@Prasun Biswas – Direct proof was indeed of great help. I don't know why I complicated things up..

Vighnesh Raut - 6 years, 1 month ago

@Vighnesh Raut – In the link, Christian Blatter's solution establishes a recursion, which I suppose is closely related to induction. One of the other solutions does use induction, so it can be done this way but it does seem to involve more work.

Brian Charlesworth - 6 years, 1 month ago

@Brian Charlesworth – yep, his solution was of great help, and when I saw the graph of the function, it was clear that rule of symmetry could be applied here...Thanks for your kind help Sir.

Vighnesh Raut - 6 years, 1 month ago

A product of our times

The answer is 4029.

1 solution

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