Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 3

A real or complex sequence $(a_n)$ is a Cauchy sequence ifFor $z_n=x_n+i{y}_n$, the complex modulus givesThus the same $N$ works for both coordinate Cauchy sequences. Conversely, if both real coordinate Cauchy sequences converge to $x$ and $y$, then $|z_n-(x+iy)|\le |x_n-x|+|y_n-y|\to 0$. This proves the needed completeness for complex numbers from the assumed completeness of the real numbers.

Now suppose ${\sum }_{n\ge 0}|z_n|$ converges. Its real partial sums form a Cauchy sequence. Given $\epsilon >0$, sufficiently late tails therefore satisfy ${\sum }_{k=n+1}^m|z_k|<\epsilon$. For $s_m={\sum }_{k=0}^m{z}_k$, the triangle inequality yieldsThe complex partial sums are a Cauchy sequence, hence converge by the coordinate argument. Therefore absolute convergence implies convergence for a complex series.

Define the radius of convergence of the power series bywith $R=\infty$ when this set is unbounded. It is nonempty because the power series converges at zero.

If $|z|<R$, the definition of supremum supplies a point $w$ of convergence with $|z|<|w|$. The terms $a_n{w}^n$ of a convergent series tend to zero and are bounded, say $|a_n{w}^n|\le M$. ThereforeThe right-hand side has a convergent geometric series, since $|z|/|w|<1$. The comparison test proves absolute convergence at $z$.

If $|z|>R$ and the power series converged at $z$, then $|z|$ would belong to the set whose supremum is $R$, an immediate contradiction. ThusNo assertion about the boundary $|z|=R$ follows from this argument. If $R=0$ the first assertion is vacuous; if $R=\infty$ the second is vacuous.

For fixed $z\ne 0$, the ratio of successive absolute values isThe ratio test says that a series converges absolutely when this ratio has a limit of a sequence strictly below one. Thus this power series converges absolutely for every complex $z$; at zero the conclusion is immediate. Its radius of convergence is $\boxed{R=\infty }$.

Split the power series into even and odd indices:whenever the two series converge absolutely. If $|z|<1$, the even part is a convergent geometric series, while the odd part is bounded termwise in absolute value by ${\sum }_j|z{|}^{2j+1}$, also a convergent geometric series. Hence there is absolute convergence throughout $|z|<1$.

If $|z|>1$, the odd terms have absolute values $|z{|}^{2j+1}/(2j+1{)}^2$. Their successive ratio tends to $|z{|}^2>1$, so these terms grow eventually and do not tend to zero. The term test for divergence therefore forces divergence of the original power series. Consequently its radius of convergence is