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

We saw in part (a) that

$$f(x)={\frac {2}{2+x}}+{\frac {2}{1-x}}=\sum _{n=0}^{\infty }\left({\frac {-x}{2}}\right)^{n}+\sum _{n=0}^{\infty }2x^{n},$$

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

$$-1<-x/2<1\iff 1>x/2>-1\iff 2>x>-2,$$

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

$$\sum _{n=0}^{\infty }x^{n}+\sum _{n=0}^{\infty }-x^{n}=0$$

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:

$$S_{N}=\sum _{n=0}^{N}\left(\left({\frac {-x}{2}}\right)^{n}+2x^{n}\right)=\sum _{n=0}^{N}{\frac {1}{2^{n}}}+\sum _{n=0}^{N}2(-1)^{n}={\frac {1-(1/2)^{N+1}}{1-(1/2)}}+2\cdot {\begin{cases}1&{\text{ if }}N{\text{ is even}}\\0&{\text{ if }}N{\text{ is odd }}\end{cases}},$$

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 $\lim _{N\to \infty }S_{N}$ 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.