Splendid Series

The Chebyshev polynomials of the second kind are defined by the recurrence relation $U_{n+1}(x) \; = \; 2xU_n(x) - U_{n-1}(x) \hspace{2cm} U_0(x) = 1\,,\, U_1(x) \; = \; 2x$ and are such that $U_n(\cos\theta) \; = \; \frac{\sin((n+1)\theta)}{\sin\theta}$ Thus the zeros of $U_n(x)$ are $\cos \left(\tfrac{k\pi}{n+1}\right)$ for $1 \le k \le n$ .

It is a simple induction that the leading term of $U_{2n}(x)$ is $4^nx^{2n}$ , and that $U_{2n}(0) = (-1)^n$ for all $n \ge 0$ . Since $a_n \; =\; \prod_{k=1}^{2n+1} \cos \left(\frac{k\pi}{2n+1}\right) \; =\; -\prod_{k=1}^{2n} \cos\left(\frac{k\pi}{2n+1}\right)$ we see that $a_n$ is minus the product of the zeros of $U_{2n}$ , and hence $a_n \; = \; - \frac{U_{2n}(0)}{4^n} \; = \; \frac{(-1)^{n+1}}{4^n}$ Thus we deduce that $S = -\tfrac45$ , and hence $\frac{1}{1+S} = \boxed{5}$ .

I, for one, am most comfortable with this method, though I'm sure there are others:

Note that $\cos{(\frac{(2n+1)\pi}{2n+1})}=\cos{\pi}=-1$ so $a_n=\cos{(\frac{\pi}{2n+1})} \cdot \cdot \cdot \cos{(\frac{n\pi}{2n+1})} \cos{(\frac{(n+1)\pi}{2n+1})} \cdot \cdot \cdot \cos{(\frac{2n\pi}{2n+1})}(-1)=-\cos{(\frac{\pi}{2n+1})} \cdot \cdot \cdot \cos{(\frac{n\pi}{2n+1})} \cos{(\frac{(n+1)\pi}{2n+1})} \cdot \cdot \cdot \cos{(\frac{2n\pi}{2n+1})}$ .

Let's consider another series $b_n=\sin{(\frac{\pi}{2n+1})} \cdot \cdot \cdot \sin{(\frac{n\pi}{2n+1})} \sin{(\frac{(n+1)n\pi}{2n+1})} \cdot \cdot \cdot\sin{(\frac{2n\pi}{2n+1})}$ . Multiplying them together gives us:

$a_nb_n=-\sin{(\frac{\pi}{2n+1})}\cos{(\frac{\pi}{2n+1})} \cdot \cdot \cdot \sin{(\frac{n\pi}{2n+1})} \cos{(\frac{n\pi}{2n+1})} \sin{(\frac{(n+1)\pi}{2n+1})} \cos{(\frac{(n+1)\pi}{2n+1})} \cdot \cdot \cdot \sin{(\frac{2n\pi}{2n+1})}\cos{(\frac{2n\pi}{2n+1})}$

$a_nb_n=-\frac{1}{2}(2\sin{(\frac{\pi}{2n+1})}\cos{(\frac{\pi}{2n+1})}) \cdot \cdot \cdot \frac{1}{2}(2\sin{(\frac{n\pi}{2n+1})} \cos{(\frac{n\pi}{2n+1})}) \frac{1}{2}(2\sin{(\frac{(n+1)\pi}{2n+1})} \cos{(\frac{(n+1)\pi}{2n+1})}) \cdot \cdot \cdot \frac{1}{2}(2\sin{(\frac{2n\pi}{2n+1})}\cos{(\frac{2n\pi}{2n+1})})$

$a_nb_n=-\frac{1}{2}\sin{(\frac{2\pi}{2n+1})} \cdot \cdot \cdot \frac{1}{2}\sin{(\frac{2n\pi}{2n+1})} \frac{1}{2}\sin{(\frac{(2n+2)\pi}{2n+1})} \cdot \cdot \cdot \frac{1}{2}\sin{(\frac{4n\pi}{2n+1})}$

$a_nb_n=-\frac{1}{2^{2n}}(\sin{(\frac{2\pi}{2n+1})} \cdot \cdot \cdot \sin{(\frac{2n\pi}{2n+1})} \sin{(\frac{(2n+2)\pi}{2n+1})} \cdot \cdot \cdot \sin{(\frac{4n\pi}{2n+1})}$

Note that $\sin{(\frac{(2n+2)\pi}{2n+1})}=-\sin{(\frac{(2n+2)\pi}{2n+1}-\pi)}=-\sin{(\frac{\pi}{2n+1})}$ and $\sin{(\frac{4n\pi}{2n+1})}=-\sin{(\frac{4n\pi}{2n+1}-\pi)}=-\sin{(\frac{(2n-1)\pi}{2n+1})}$ .

Therefore:

$a_nb_n=-\frac{1}{2^{2n}} (-\sin{(\frac{\pi}{2n+1})}) (\sin{(\frac{2\pi}{2n+1})} \cdot \cdot \cdot (-\sin{(\frac{(2n-1)\pi}{2n+1})}) \sin{(\frac{2n\pi}{2n+1})}$

$a_nb_n=-\frac{1}{2^{2n}} (-1)^{n} \sin{(\frac{\pi}{2n+1})} \sin{(\frac{2\pi}{2n+1})} \cdot \cdot \cdot \sin{(\frac{(2n-1)\pi}{2n+1})} \sin{(\frac{2n\pi}{2n+1})}$

$a_nb_n=-\frac{1}{2^{2n}} (-1)^{n} b_n$

$a_n=-\frac{1}{2^{2n}} (-1)^{n}$

$a_n=-\frac{(-1)^{n}}{(2^2)^n}$

$a_n=-(\frac{-1}{2^2})^n$

$a_n=-(-\frac{1}{4})^n$

So the initial term $a_0=-(-\frac{1}{4})^0=-1$ and the common ratio is $-\frac{1}{4}$ . Therefore:

$S=\frac{-1}{1-(-\frac{1}{4})}=\frac{-1}{1+\frac{1}{4}}$ and $1+S=1+\frac{-1}{1+\frac{1}{4}}=\frac{\frac{1}{4}}{1+\frac{1}{4}}$

Then $\frac{1}{1+S}=\frac{1+\frac{1}{4}}{\frac{1}{4}}=4+1=\boxed{5}$