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Calculus Level 5

$\large \lim_{n \to \infty}\sum\limits_{m=1}^{n} \prod\limits_{k=1}^{m} \cos^{2} \! \left(\frac{π(2k-1)}{2(2m+1)}\right)=\frac{A}{B}$

Find $A+B$ where $A$ and $B$ are coprime positive integers.

Inspired by Otto's problem .

The answer is 20.

1 solution

Otto Bretscher
Sep 3, 2015

What a beautiful and fascinating problem! Thank you!

I worked out a kind of rambling and roundabout solution, late at night... there has to be a better way. Along the way we will use the symmetries $\cos(\pi-t)=-\cos(t)$ and $\sin(\pi-t)=\sin(t)$ as well as the formulas for $\cos^2(t)$ and $\sin^2(t)$ in terms of $\cos(2t)$ . All products will be taken form $k=1$ to $k=m$ unless stated otherwise. Let $t=\frac{\pi}{2m+1}$ . Then the given product is

$\frac{1}{2^m}\prod(1+\cos((2k-1)t)=\frac{1}{2^m}\prod(1-\cos(2kt))$ $=\prod\sin^2(kt)=\prod_{k=1}^{2m}\sin(kt)=\frac{2m+1}{4^m}.$

The last formula has been discussed here .

Now $\sum_{m=1}^{\infty}\frac{2m+1}{4^m}=\frac{11}{9}$ , an arithmetic-geometric series, so that $a+b=\boxed{20}$

I'm sorry to report that it is "back to work" for me... classes start after the US "Labour Day".. with a heavy heart I need to say goodbye to Brilliant, but I will be back in December, if all goes well. Thank you for all the fun!

Moderator note:

Aw, we will miss you! Come back in December with a bunch of interesting problems!

You can send your students here for extra credit :)

There is a much easier way.

We can deduce $z^{2n}+1=\prod\limits_{k=1}^{n}\left(z^2-2z\cos\left(\frac{(2k-1)\pi}{2n}\right)+1\right)$

Using $z^{2n}+1=(z^2+1)(1-z^2+z^4-...+z^{2n-2})$ (when n is odd) and plugging $z=i$ . The product written above can be worked out from that by considering odd n.

I'll miss your wonderful problems and solutions Otto.

Isaac Buckley - 5 years, 9 months ago

Very elegant indeed! Thanks.

Otto Bretscher - 5 years, 9 months ago