Review : Power Series — Question 2

Question 2 – Solution

Strategy. Treat the missing coefficients explicitly. A convergent positive comparison series controls the interior, while selected large terms control boundary growth.

Step 1: Identify the coefficients. We have $a_n=n$ when $n=2^k$ for an integer $k\ge 0$, and $a_n=0$ otherwise, including $n=0$. Consecutive coefficient ratios repeatedly have zero denominators, so the usual coefficient-ratio shortcut is unavailable. This does not prevent other convergence tests from determining the radius.

Step 2: Establish the radius directly. For $r=|x|<1$, $$\sum_{k=0}^{\infty}2^kr^{2^k}\le\sum_{n=1}^{\infty}nr^n =\frac{r}{(1-r)^2}<\infty.$$ Thus the series converges absolutely throughout $|x|<1$. If $|x|>1$, the magnitudes $2^k|x|^{2^k}$ do not tend to zero; indeed they grow without bound. Therefore $\boxed{R=1}$. The comparison sum follows by differentiating the geometric series inside its radius.

Step 3: Test parity at the endpoints. At $x=1$, the terms are $2^k$ and fail the term test. At $x=-1$, the first term is $-1$, but every term with $k\ge 1$ is $+2^k$, since $2^k$ is even. There is no alternating tail. Both endpoint series diverge, so $\boxed{I=(-1,1)}$, with absolute convergence at every point of this interval.

Step 4: Prove unbounded growth despite the gaps. For $0<x_m<1$, every term is positive. Keeping just the term indexed by $k=m$ gives $$\boxed{L(x_m)\ge 2^m x_m^{2^m}=2^m/e\longrightarrow\infty.}$$ Also $x_m\uparrow 1$. Thus $L$ is not bounded near $1$. In fact the sum is increasing on $[0,1)$ because each term is increasing, so this sequence of lower bounds proves $L(x)\to+\infty$ as $x\to 1^-$. Sparse powers do not offset the growth of the coefficients and their near-boundary terms.