# 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?

• That's not a power series. – zhw. Jun 4 '15 at 20:47
• I thought the same, but I figured that they are asking for convergence .... and we know that if a power series convergence on some interval, then its sum function is actually continuous, and that, I thought, would explain the vague question of "show convergence". – Michael Maxi Jun 4 '15 at 20:49
• It is a power series in $2x/(1+x^2)$. And inasmuch as $a_n$ is bounded and $|x|<1$, which implies that $|y|=|2x/(1+x^2)|<1$, the series converges. – Mark Viola Jun 4 '15 at 20:49
• Did you mean to sum over $k$, not $n$? – Barry Jun 4 '15 at 20:51

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.

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 $-1 < x < 1$.