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