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} $


$$\sum_{1\le r\le n} \frac1{2^r}\tan \frac x{2^r}=\frac 1{2^{n+1}}\cot \frac x{2^{n+1}}-\cot x$$

Now, putting $\frac1{2^{n+1}}=h,$ $$\lim_{n\to\infty}\frac 1{2^{n+1}}\cot \frac x{2^{n+1}}=\lim_{h\to0}\cos hx\cdot\lim_{h\to0}\frac{hx}{\sin hx}\cdot \frac1x=\frac1x$$

  • $\begingroup$ Hmm... is my answer right? And I am more interested in a source for this question than alternative methods. I surely appreciate your effort though. Thank you :) $\endgroup$ – Isomorphism May 15 '13 at 12:01
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    $\begingroup$ @Isomorphism, yes our answers concur $\endgroup$ – lab bhattacharjee May 15 '13 at 12:09
  • $\begingroup$ Thanks for the detailed answer too :) Have you seen this question before? $\endgroup$ – Isomorphism May 15 '13 at 12:10
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    $\begingroup$ @Isomorphism, just re-discovered Ex 2 in Page:414 of archive.org/details/planetrigonomet00lonegoog . There are many such before/after this $\endgroup$ – lab bhattacharjee May 15 '13 at 14:46

The identity $$ \cot(x)+\frac12\tan(x/2)=\frac12\cot(x/2) $$ sort of leads you to the solution. Iterating gives your sum.

  • $\begingroup$ I didn't know this one. $\endgroup$ – Soham Chowdhury May 15 '13 at 13:38
  • $\begingroup$ @SohamChowdhury: the proof is given in lab's answer, but I reformat it here: $$ \begin{align} \cot(x)-\tan(x) &=\frac{\cos(x)}{\sin(x)}-\frac{\sin(x)}{\cos(x)}\\ &=2\frac{\cos^2(x)-\sin^2(x)}{2\sin(x)\cos(x)}\\ &=2\frac{\cos(2x)}{\sin(2x)}\\ &=2\cot(2x) \end{align} $$ the rest is repackaging to make it more obvious how to get the recursion. $\endgroup$ – robjohn May 15 '13 at 14:27

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