Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 18 (b)/Solution 1

We saw in part (a) that $$f(x)=\frac{2}{2+x}+\frac{2}{1-x}=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{-x}{2}\right)}^n+\sum _{n=0}^{\mathrm{\infty }}2x^n,$$ where the first series converges if and only if $-1<-x/2<1$, and the second series converges if and only if $-1<x<1$. Since have the following sequence of equivalences $$-1<-x/2<1⟺1>x/2>-1⟺2>x>-2,$$ we see that the first series has radius of convergence $2$.

We see then that, if $-1<x<1$, then both series converge, and so their sum converges as well. This suggest that the radius of convergence is $R=1,$ but we still need to check that the radius of convergence is not larger. Indeed, the example $$\sum _{n=0}^{\mathrm{\infty }}{x}^n+\sum _{n=0}^{\mathrm{\infty }}-x^n=0$$ shows that it is possible a sum of series to converge with a larger radius than the radii of the summand series. To see that this does not happen for us, we must check how the series for $f(x)$ behaves at $x=\pm 1$.

At $x=-1$, we find the partial sum: $$S_N=\sum _{n=0}^N({\left(\frac{-x}{2}\right)}^n+2x^n)=\sum _{n=0}^N\frac{1}{2^n}+\sum _{n=0}^{N}2(-1{)}^n=\frac{1-(1/2{)}^{N+1}}{1-(1/2)}+2\cdot \{\begin{array}{ll}1& \text{ if }N\text{ is even}\\ 0& \text{ if }N\text{ is odd }\end{array},$$ which may be computed using the formula $\sum _{n=0}^N{x}^n=\frac{1-x^{N+1}}{1-x}$. The point is that, because of the second term in $S_N$, the limit $\underset{N\to \mathrm{\infty }}{\lim }{S}_N$ does not exist, so our series for $f(x)$ does not converge for $x=-1$.

At $x=1$, the $n^{\text{th}}$ summand of the series for $f(x)$ is ${\left(\frac{-x}{2}\right)}^n+2x^n={\left(\frac{-1}{2}\right)}^n+2,$ which is positive and does not converge to 0 as $n$ tends to infinity. Therefore the series does not converge at $x=1$ either. This completes the proof that the radius of convergence is indeed 1.