Is this a power series? Is the following a power series?
$$\sum_{n=0}^\infty a_k \left( \frac{2x}{1+x^2} \right)^k \ , x \in (-1,1)$$
where $a_k$ is a bounded sequence. 
I was asked to show that this power series converges, and that the sum function was continuous. 
Two things confused me: Why are they calling this weird-looking thing a "power series"? And why are they asking for "convergence"? Isn't it customary to ask for either pointwise or uniform convergence, especially with the question of continuous sum function in mind?
 A: It's a power series in $u = \frac{2x}{1+x^2}$.
As such, if $|u| < 1$ for $x \in (-1, 1)$, then as long as $a_k$ is bounded, the series is convergent.
Fortunately, we can indeed show that $|u| < 1$.  When $0 \leq x < 1$,
$$
0 < (x-1)^2 \leq 1
$$
$$
0 < x^2-2x+1 \leq 1
$$
$$
-1-x^2 < -2x < -x^2
$$
$$
1+x^2 > 2x > x^2 \geq 0
$$
$$
0 \leq \frac{2x}{1+x^2} < 1
$$
Complete similar reasoning for $-1 < x \leq 0$, and you are done.
A: It hasn't been written as a power series in $x$, but it can be expressed as one.
Consider the term $\frac{2x}{1+x^2}$. If we expand the geometric series here, we find
$$
\frac{2x}{1 + x^2} = 2x(1 - x^2 + x^4 - x^6 + \cdots) = 2x - 2x^3 + 2x^5 - 2x^7 + \cdots
$$
which when raised to the $k$-th power gives you something of the form
$$
2^kx^k + O(x^{k+1})
$$
As such, you can expand this out and you will only find finitely many terms of that contribute to each coefficient of a power series in $x$; that is only the first $k$ terms in the expression will contribute to any coefficient $b_k$ of the resulting series $\sum_k b_kx^k$, and so it is well-defined---at least for $-1 < x < 1$.
