Can the following summation be written in a finite number of terms:


I tried to simplify the expression using trigonometric identities and then converting the infinite summation into a definite integral. I couldn't think of any way to get it into that since the variable $r$ occurs as an exponent.

So is there any other possible simplification for the expression?


1 Answer 1


Let $a_n = \dfrac{\tan(\frac{x}{2^n})}{2^{n-1}\cos(x/2^{n-1})}$. We want to find $\sum_{n=1}^{\infty} a_n$.

Let's write out the first term: $$a_1 = \frac{\tan(x/2)}{\cos(x)}$$

and invoke the identity $\tan(\frac{x}{2}) = \csc(x)-\cot(x)$:

$$a_1 = \frac{\csc(x)-\cot(x)}{\cos(x)} \\ = \frac{1}{\sin(x)\cos(x)} - \frac{\cos(x)}{\sin(x)\cos(x)} \\ = \frac{2}{2\sin(x)\cos(x)} - \frac{1}{\sin(x)} \\ = \frac{2}{\sin(2x)} - \csc(x) \\ a_1 = 2\csc(2x) - \csc(x).$$

By performing the same procedure for the next couple terms ($a_2$ and $a_3$), we have the following:

$$a_2 = \csc(x) - \frac{1}{2}\csc(x/2) \\ a_3 = \frac{1}{2}\csc(x/2) - \frac{1}{4}\csc(x/4) \\ \vdots \\ \vdots \\ a_n = \frac{1}{2^{n-2}} \csc(x/2^{n-2}) - \frac{1}{2^{n-1}}\csc(x/2^{n-1}).$$

Now it's time to consider partial sums. Let $s_n = a_1 + a_2 + \cdots + a_n$. A pattern emerges quickly because of the telescoping nature of the terms:

$$s_1 = a_1 = 2\csc(2x) - \csc(x) \\ \ \\ \ \\ s_2 = s_1 + a_2 = [2\csc(2x) - \csc(x)] + [\csc(x) - \frac{1}{2}\csc(x/2)] \\ s_2 = 2\csc(2x) - \frac{1}{2}\csc(x/2) \\ \ \\ \ \\ s_3 = s_2 + a_3 = [2\csc(2x) - \frac{1}{2}\csc(x/2)] + [\frac{1}{2}\csc(x/2) - \frac{1}{4}\csc(x/4)] \\ s_3 = 2\csc(2x) - \frac{1}{4}\csc(x/4) \\ \vdots \\ \vdots \\ s_n = 2\csc(2x) - \frac{1}{2^{n-1}}\csc(x/2^{n-1}).$$

Now that we have an expression for the $n$-th partial sum, we can attempt considering the limiting case since $\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} s_n$. That is,

$$\lim_{n \to \infty} s_n \\ = \lim_{n \to \infty} \left[ 2\csc(2x) - \frac{1}{2^{n-1}}\csc(x/2^{n-1}) \right] \\ = 2\csc(2x) - \lim_{n \to \infty} \left[\frac{1}{2^{n-1}}\csc(x/2^{n-1}) \right] \\ = 2\csc(2x) - \lim_{n \to \infty} \left[\frac{1}{2^{n-1}\sin(x/2^{n-1})}\right] $$

Unfortunately, this limit is not defined. However, for $x\neq 0$, it can be shown that we're left with

$$\lim_{n \to \infty} s_n = 2\csc(2x) - \frac{1}{x}.$$

So altogether, and replacing $n$ with $r$ and replacing $x$ with $\theta$, we have $$ \sum_{r=1}^{\infty} \frac{\tan(\frac{\theta}{2^r})}{2^{r-1}\cos(\theta/2^{r-1})} = 2\csc(2\theta) - \frac{1}{\theta}.$$

I'm not sure how to find out the restrictions on $x$ this late at night. Fro example, since $x=0$ gives zero for the sum but not this answer, perhaps someone else could take over at this point. However, I hope the derivation is of some use to you!

  • $\begingroup$ Thanks that helps. But what's telescoping nature of a series? $\endgroup$
    – Piyush
    Dec 20, 2015 at 12:18
  • $\begingroup$ Notice that the middle terms of $a_1 +a_2$ cancel out. Indeed, the middle terms of $a_1 + a_2 + a_3$ cancel out. You're just left with something from the very first term and then something from the very last term in any given partial sum. Read up on it here: en.wikipedia.org/wiki/Telescoping_series $\endgroup$
    – Xoque55
    Dec 20, 2015 at 16:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .