Why is the power series of $\tan(x)$ not convergent everywhere $\cos(x)$ is non-zero? The Taylor series expansion of $\tan(x)$ centered at $0$ has a radius of convergence of $\pi/2$, which means the power series will not converge for $|x|>\pi/2$.
How can this be when you consider $\tan(x)=\sin(x)/\cos(x)$, so Taylor series of $\tan(x)$ is just the Taylor series of $\sin(x)$ divided by the power series of $\cos(x)$, both of which converge everywhere.
At some point $|x|>\pi/2$ where $\cos(x)$ is not equal to $0$, how can the power series of $\tan(x)$ diverge? It is simply the quotient of two convergent series at that point which seems to me like it shouldn't have any problems.
 A: The representation of a function may or may not provide all the values of where a function is definable. There are whole books dedicated to writing functions in multiple different ways in order to understand the values of the function on different subsets of the complex plane (integrals, dirichlet series, power series, etc.). Take for example, the Riemann-Zeta function; this function has many many different representations, most of which only exist on some half plane of $\mathbb{C}$. Yet it is known that the Riemann-Zeta function is analytic everywhere in the complex plane and has only one simple pole at 1; a non-obvious fact if you were to look at any of the representations of the Riemann-Zeta function. The main points are:

*

*A representation may or may not provide all the values of where a function is analytic.


*A function can have many representations.


*Representations have their own properties apart from the functions they might represent.
Power series are one kind of representation for analytic functions. Power series enjoy the property of uniqueness. Some functions are only really known by their power series expansion and have no "closed form". Power series expansions have what we call a radius of convergence, this radius is what defines where a power series is useful and tells us values of the underlying function. The radius of convergence is always as large as possible for the underlying function represented. If you stray outside that radius of convergence then the power series is unhelpful and does not tell you the proper values of the underlying function that is being represented. Take for example, $$f(z) = \frac{1}{1-z}$$ this function is clearly defined and analytic in all of $\mathbb{C}-\{1\}$. However, "a" representation of $f(z)$ is the power series centered at zero, namely for every $z\in B(0,1)$, we have, $$f(z) = \sum_{k=0}^\infty z^k$$ This power series is only useful on $B(0,1)$ and is not useful outside that set. The power series agrees with $f(z)$ wherever it converges, but the power series is not guaranteed to converge wherever $f(z)$ exists. This is true of any representation of a function. We could have picked another power series instead, say one centered at -1, $$f(z)=\sum_{k=0}^\infty \frac{1}{2^{k+1}}(z+1)^k$$ This power series converges with radius 2 instead of radius 1. Secretly, I know $f(z)$ has a pole at 1, and since power series always maximize their radius then I knew the radius had to be 2 for this power series because -1 is 2 units away from 1, even before I calculated the coefficients. Likewise, if I found the power series at $-i$, I already know the radius would be $\sqrt{2}$, since that is the distance from $-i$ to the pole at 1 for $f(z)$.
The power series of $\tan(z)$ at zero has radius $\frac{\pi}{2}$ because $\tan(z)$ has at least one singularity on the circle $B(0,\frac{\pi}{2})$. If it didn't have a singularity somewhere on that circle, then the radius of convergence would have been bigger. The radius of convergence always maximizes where the underlying function is analytic.
A: The reason is that, if a function $f:\Bbb C\to \Bbb C$ has a power-series representation $$f(z)=\sum_{k=0}^\infty a_kx^k$$ that fails to converge at a point $z=c\in\Bbb C$, then it cannot converge at any point $z$ such that $|z|>|c|$. This is because, roughly speaking, if the terms $a_nc^n$ do not tend to zero fast enough for the series to converge as $n\to\infty$, then $a_nz^n$ will tend to zero even more slowly when $|z|>|c|$, or not even tend to zero at all, and so a fortiori will prevent the series from converging.
In the case of the tangent function, because it blows up at $\frac12\pi$, the Taylor series $\tan z=\sum_{k=0}^\infty a_kx^k$ for it will not converge at $z=\frac12\pi$ and so will not converge outside the circle $|z|=\frac12\pi$.
The borderline case is when $|z|$ equals the radius of convergence. In some instances, the series will converge (conditionally) at certain points. For example, $$\ln(1-z)=-\sum_{k=1}^\infty\frac1kz^k$$still converges at $z=-1$.
A: While $\tan(z)=\frac{\sin(z)}{\cos(z)}$, it is not true that the Taylor series of $\tan$ is just the Taylor series of $\sin$ divided by the Taylor series of $\cos$! I can confidently say this even though I'm not even sure what you mean be dividing the two series. I can imagine two ways to do this, none of which result in the Taylor series of $\tan$.
1: Dividing term by term
This breaks essentially immediately, since $\sin$ is an odd function, while $\cos$ is an even function. Being even, the Taylor series of $\cos$ has vanishing coefficients at odd powers. Being odd, the Taylor series of $\sin$ has non-vanishing coefficients at odd powers. So we'd be dividing something non-zero by zero. So this clearly doesn't work at all.
2: Dividing the partial sums
Series are essentially just sequences. The series $\sum_{k=0}^\infty a_k$ is simply the sequence $b_n:=\sum_{k=0}^n a_k$, and the series converges to the limit of this sequence. So we could take two power series $\sum_{k=0}^\infty a_k z^k$ and $\sum_{k=0}^\infty b_k z^k$ and consider the sequence
$$c_n:=\frac{\sum_{k=0}^n a_k z^k}{\sum_{k=0}^n b_kz^k}.$$
In fact, wherever the denominator doesn't converge to $0$, this sequence converges, and it converges to
$$\frac{\sum_{k=0}^\infty a_k z^k}{\sum_{k=0}^\infty b_kz^k},$$
so if we divide the Taylor series of $\sin$ and $\cos$ this way, the sequence will, in fact, converge to $\frac{\sin(z)}{\cos(z)}$ at all points where this quotient exists. But! This sequence is not a power series. Power series are sequences of the form $b_n=\sum_{k=0}^n a_k (z-z_0)^k$. The above quotient is not of this form, and it cannot be transformed to take that form. The Taylor series of $\tan$ is not equal to this sequence of quotients.
