Review : Power Series — Question 1

Question 1 – Solution

Strategy. Normalize the displacement from the center, test the endpoints as numerical series, then justify any boundary value independently.

Step 1: Apply the ratio test away from the center. Write $z=(x-2)/3$. For $z\ne 0$, the ratio of successive term magnitudes is $|z|n/(n+1)\to|z|$. Thus the center is $2$ and $\boxed{R=3}$. The series converges absolutely for $|z|<1$ and diverges for $|z|>1$. At the center all terms vanish. The ratio test is inconclusive at $z=\pm 1$.

Step 2: Resolve the two endpoints. At $x=-1$, the series is $\sum_{n\ge 1}(-1)^n/n$, which converges by alternation but not absolutely. At $x=5$, it is the divergent harmonic series. Therefore $$\boxed{I=[-1,5),\quad\text{absolute on }(-1,5),\quad \text{conditional only at }-1.}$$ A radius specifies distances, not whether either boundary point is included.

Step 3: Integrate the geometric series. On any compact subinterval of $(-1,1)$, the geometric series converges uniformly, so integration from $0$ to $z$ gives $$\sum_{n=1}^{\infty}\frac{z^n}{n} =\int_0^z\frac{du}{1-u}=-\log(1-z).$$ This also holds for negative $z$ with the usual oriented integral. Both sides vanish at $z=0$, fixing the constant. Hence $\boxed{F(x)=-\log(1-(x-2)/3)}$ for $-1<x<5$.

Step 4: Establish the boundary value. The finite geometric identity gives $\sum_{k=0}^{N-1}(-u)^k=(1-(-u)^N)/(1+u)$. Integrating over $[0,1]$, its difference from $\log 2$ has magnitude at most $\int_0^1u^N\,du=1/(N+1)\to 0$. Negating the resulting alternating harmonic sum proves $\boxed{F(-1)=-\log 2}$. At $x=5$ there is no alternation and harmonic divergence persists; the interior logarithm also tends to $+\infty$ there.