How to find the limit of a sequence? Question:
If $0 < x < \frac{\pi}{2}$ and $f_k(x) = \tan(x)+\frac{1}{2}\tan(x/2)+ ...+\frac{1}{2^k}\tan(x/2^k)$.
In Sigma Notation:
$$f_k(x) = \sum_{n=0}^k \frac{1}{2^n}\tan\frac{x}{2^n}$$
Find $$\lim_{k\to \infty}f_k(x)$$
I'm not quite sure how to take the limit of a sequence, I've never seen this before, If someone could help me solve this it would be greatly appreciated.
 A: Just observe that $$\tan\alpha=\cot\alpha-2\cot 2\alpha,$$ and you'll get
$$S_k=\sum_{n=0}^k \frac{1}{2^n}\tan\frac{\alpha}{2^n}=\tan\alpha+\sum_{n=1}^k\frac{1}{2^n}\cot\frac{\alpha}{2^n}-\sum_{n=1}^k\frac{1}{2^{n-1}}\cot\frac{\alpha}{2^{n-1}}=$$
$$=\tan\alpha-\cot\alpha+\frac{1}{2^k}\cot\frac{\alpha}{2^k}.$$
Since $x\cot\alpha x\to \frac{1}{\alpha}$ when $x\to 0$, the final result is $$\lim_{k\to\infty} S_k=\tan\alpha-\cot\alpha+\frac{1}{\alpha}=\frac{1}{\alpha}-2\cot2\alpha.$$
A: I assuming that is $\lim_{k\rightarrow\infty}f_{k}\left(x\right)=\sum_{n\geq0}\frac{\tan\left(x/2^{n}\right)}{2^{n}}.
 $ Using the Taylor series of tangent and cotangent we have $$\sum_{n\geq0}\frac{\tan\left(x/2^{n}\right)}{2^{n}}=\sum_{n\geq0}\frac{1}{2^{n}}\sum_{m\geq1}\frac{B_{2m}\left(-1\right)^{m}2^{2m}\left(1-2^{2m}\right)}{\left(2m\right)!}\frac{x^{2m-1}}{2^{2mn-n}}=\sum_{m\geq1}\frac{B_{2m}\left(-1\right)^{m}2^{2m}\left(1-2^{2m}\right)}{\left(2m\right)!}x^{2m-1}\frac{2^{2m}}{2^{2m}-1}=
 $$ $$=-2\sum_{m\geq1}\frac{B_{2m}\left(-1\right)^{m}2^{2m}}{\left(2m\right)!}\left(2x\right)^{2m-1}=\frac{1}{x}-2\cot\left(2x\right).
 $$
