Review : Power Series — Question 8

Question 8 – Solution

Strategy. Apply the geometric test to the substituted variable first; then derive and test a separate series in powers of $x$.

Step 1: Pull back the geometric convergence condition. The displayed series converges exactly when $|(x^2-4)/2|<1$, or $2<x^2<6$. Thus its convergence set is $$\boxed{(-\sqrt 6,-\sqrt 2)\ \cup\ (\sqrt 2,\sqrt 6).}$$ At $x=\pm\sqrt 2$ the ratio is $-1$; at $x=\pm\sqrt 6$ it is $1$. In each case the terms fail to approach zero. Elsewhere outside the two intervals the ratio has magnitude greater than one. These polynomial terms do not have the ordinary form $a_n(x-c)^n$ for a fixed center.

Step 2: Sum on the justified set. Geometric summation gives $\boxed{P(x)=2/(6-x^2)}$ on the two intervals above. This rational formula is defined for every real $x\ne\pm\sqrt 6$, which is a much larger set. Its being finite at other points says nothing by itself about convergence of the displayed polynomial series there.

Step 3: Construct a distinct centered power series. Factoring the denominator differently gives $$\boxed{\frac 2{6-x^2}=\frac{1/3}{1-x^2/6} =\sum_{n=0}^{\infty}\frac{x^{2n}}{3\cdot 6^n}},\qquad |x|<\sqrt 6.$$ Its odd coefficients vanish. Its radius is $\sqrt 6$ and both endpoints diverge because the terms there equal $1/3$. The new convergence interval is $(-\sqrt 6,\sqrt 6)$.

Step 4: Compare actual sums, not just formulas. At $x=0$, the original terms are $(-2)^n$, so the original series diverges; the new series sums to $\boxed{1/3}$. At $x=1$, the original ratio is $-3/2$ and its series diverges, while the new series sums to $\boxed{2/5}$. On $\sqrt 2<|x|<\sqrt 6$, both converge and both equal $2/(6-x^2)$. The new representation adds valid points; it does not retroactively change the convergence set of the original sequence of partial sums.