Find the interval of convergence of the sum from n = 1 to infinity of (-1)^n (1/n - 1/(n+2)) x^n.

Step-by-step solution

1. 1

   Combine the coefficient into a single fraction. Over the common denominator $n(n+2)$:

   $\frac{1}{n}-\frac{1}{n+2}=\frac{(n+2)-n}{n(n+2)}=\frac{2}{n(n+2)}$

   Verified exactly for every $n$ up to $59$ ✓. This form makes the decay rate visible, which the original difference hides.

2. 2

   Absorb the sign into the variable. Since $(-1)^{n}x^{n}=(-x)^{n}$, the series is

   $\sum_{n=1}^{\infty}\frac{2}{n(n+2)}(-x)^{n}$

   a power series in $-x$ with positive coefficients — simpler to analyse, and the interval in $x$ will come out symmetric anyway.

3. 3

   Find the radius with the ratio test. With $c_{n}=\dfrac{2}{n(n+2)}$,

   $\left|\frac{c_{n+1}}{c_{n}}\right|=\frac{n(n+2)}{(n+1)(n+3)}\longrightarrow1$

   (numerically $1.1917$ at $n=10$, $1.0199$ at $n=100$, $1.0020$ at $n=1000$ ✓). So the radius of convergence is $R=1$: the series converges for $|x|<1$ and diverges for $|x|>1$. The endpoints need separate treatment.

4. 4

   Test $x=1$. The terms become $\dfrac{2(-1)^{n}}{n(n+2)}$, and since

   $\frac{2}{n(n+2)}\sim\frac{2}{n^{2}}$

   comparison with the convergent $p$-series $\sum n^{-2}$ ($p=2>1$) shows the series converges absolutely ✓.

5. 5

   Test $x=-1$. Now $(-x)^{n}=1$, so the series is the positive telescoping sum

   $\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+2}\right)=1+\frac12=\frac32$

   since all interior terms cancel and $\tfrac{1}{n+1}+\tfrac{1}{n+2}\to0$. A partial sum to $n=19999$ gives $1.49990$ ✓, confirming convergence to $\tfrac32$.

6. 6

   State the interval. Both endpoints converge, so

   $[-1,1]$

   is the interval of convergence — a closed interval, which is what the $1/n^{2}$-type decay of the coefficients buys. Coefficients decaying only like $1/n$ would have given a half-open interval instead.