Question:
If $x \neq 0$, then prove that $\displaystyle \sum_{n=1}^{\infty}\dfrac1{2^n} \tan\left(\dfrac{x}{2^n}\right) = \dfrac1{x} - \cot x.$
My answer:
I proved this result by using the following identity:
$$ \prod_{k=1}^n \cos\left(\frac{x}{2^k}\right) = \frac{\sin x}{2^n\sin \frac{x}{2^n}}$$
I took natural log on both sides of the above equation and then differentiated both sides to get
$$\sum_{k=1}^n \frac1{2^k} \tan\left(\frac{x}{2^k}\right) = \frac{1}{2^n}\cot \left(\frac{x}{2^n}\right) - \cot x.$$
Now taking the limit of $n$ to $\infty$ I get the required identity. $\blacksquare$
I have never seen the above identity before in my life. I was amused and surprised by this identity. That's because I have never seen an infinite trig series summing up to a rational function like $\dfrac1{x}$ before.
So my questions are the following:
A) First of all, is my derivation correct? Secondly, does anyone know a source for this problem?
B) Are there other 'elementary 'derivations that have an infinite trig series on one side and a rational function on the other? [I say 'elementary' to avoid Fourier series. I am guessing Fourier series must be full of such results.]
HINT: Use $$\cot y-\tan y=\frac{\cos^2yx-\sin^2y}{\sin y\cos y}=2\frac{\cos2y}{2\sin y\cos y}=2\frac{\cos2y}{\sin2y}=2\cot2y$$
$$\iff \tan y=\cot y-2\cot2y $$
Putting $y=\frac x2,\frac x{2^2},\frac x{2^3}\cdots$
$\tan \frac x2=\cot \frac x2-2\cot x $
$\tan \frac x{2^2}=\cot \frac x{2^2}-2\cot \frac x2 $
$\tan \frac x{2^3}=\cot \frac x{2^3}-2\cot \frac x{2^2} $
etc
$$\sum_{1\le r\le n} \frac1{2^r}\tan \frac x{2^r}=\frac 1{2^{n+1}}\cot \frac x{2^{n+1}}-\cot x$$
Now, putting $\frac1{2^{n+1}}=h,$ $$\lim_{n\to\infty}\frac 1{2^{n+1}}\cot \frac x{2^{n+1}}=\lim_{h\to0}\cos hx\cdot\lim_{h\to0}\frac{hx}{\sin hx}\cdot \frac1x=\frac1x$$