$$\sum_{n=1}^{\infty} \dfrac {1}{2^n}\tan\left(\dfrac{x}{2^n}\right) = \dfrac 1x - \cot x$$

I used an identity $\tan(1/2x) = \text{cot }\left(\dfrac{1}{2x}\right) - 2\text{cot}(x)$ and input $x= \dfrac{X}{2^n-1}$ into the summation.

I get a telescoping sum. When I look at the telescoping sum, I assume that all other terms cancel out except $-\cot x$. But that wasn't the right answer it's $\dfrac 1x- \text{cot}\,x$. Why? I know they looked at the telescoping sum and took the limit of the nth term of the telescoping sum as $n\to\infty$ which became $\dfrac 1x$ and added the $-\text{cot }x$. Logically to me this doesn't make sense. The sum reduces (the way I remembered (if I'am correct) $$\sum_{n=1}^{\infty} (\text{cot}\left(\dfrac{1}{2^n}\right) - \text{cot}\,\left(\dfrac{1}{2^n-1}\right)).$$

$2n$ is ahead of $2^{n-1}$. Shouldn't all there terms cancel all the way to infinity out except $-\text{cot }x$. But instead, we take the sum of the telescoping sum to the nth term and take the limit? That doesn't add up all the terms. Please help.

  • $\begingroup$ Please learn how to format on this site. Unformatted math is very hard to read (does "1/2^n-1" mean $\frac 1{2^{n-1}}$ or $\frac 1{2^n} - 1$?). A good tutorial can be found here $\endgroup$ – lulu Aug 18 '16 at 17:02
  • $\begingroup$ The first. (1/2^(n-1)) $\endgroup$ – Math Aug 18 '16 at 17:02
  • $\begingroup$ Please edit your question accordingly, using the formatting language. $\endgroup$ – lulu Aug 18 '16 at 17:03
  • $\begingroup$ Ok, hang on.... $\endgroup$ – Math Aug 18 '16 at 17:05

This is just the definition of an infinite sum. By definition, the notation $\sum_{n=1}^\infty a_n$ refers to the limit of the partial sums: $$\sum_{n=1}^\infty a_n=\lim_{N\to\infty}\sum_{n=1}^Na_n.$$

Here's an example of why intuitively we should take this definition rather than let you naively telescope to infinity like you want to. Consider the sum $$S=\sum_{n=1}^\infty 0.$$ Obviously, $S$ should be $0$: every term of $S$ is $0$. And indeed, each partial sum is $0$, so the limit of the partial sums is $0$. But here's a clever way to rewrite this sum: $$S=\sum_{n=1}^\infty (1-1).$$ This is the exact same sum, since $1-1=0$. But now we can view it as a telescoping sum, where the $-1$ in each term cancels the $1$ in the following term. When you cancel them all out, you're just left with the initial $1$. So by your logic, $S$ should be $1$.

Or, we could rewrite $S$ as $$S=\sum_{n=1}^\infty (2-2).$$ Now by telescoping, you get that $S=2$. In fact, by rewriting it in different ways and telescoping, you could claim that $S$ has any value at all.


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