Complex Power Series and Analytic Functions — Examples

Rewrite $1-z=(1-b)(1-(z-b)\mathrm{/}(1-b))$; since $b\mathrm{\ne }1$, the finite geometric identity gives the displayed infinite series when $\mathrm{\mid }(z-b)\mathrm{/}(1-b)\mathrm{\mid }<1$.

The coefficient modulus is $\mathrm{\mid }1-b{\mathrm{\mid }}^{-k-1}$, whose root limsup is $\mathrm{\mid }1-b{\mathrm{\mid }}^{-1}$, so [L2] gives radius $\mathrm{\mid }1-b\mathrm{\mid }$.

The coefficients agree with the interior re-expansion guaranteed by [L1], and no assertion is made on its boundary circle.

Regard the real coefficients $(-1{)}^{n+1}\mathrm{/}n$ as complex coefficients; by [L1] their series has sum $\log 2$.

Apply [L2] to obtain the asserted angular limit. The claim concerns only the boundary limit and introduces no logarithm branch inside the disc.

If $z\mathrm{\ne }1$, the finite identity $(1-z){\sum }_{k=1}^N{z}^k=z-z^{N+1}$ gives $\mathrm{\mid }{\sum }_{k=1}^N{z}^k\mathrm{\mid }\le 2\mathrm{/}\mathrm{\mid }1-z\mathrm{\mid }$ by [L4].

Apply [L1] to the bounded partial sums in step 1.1 and the decreasing weights $1\mathrm{/}n$. The tail is bounded by a fixed multiple of $1\mathrm{/}p$, which tends to $0$ by [L3], so the series converges.

At $z=1$ the series is the divergent $p$-series with $p=1$ by [L2]. At every other point on the circle, $\mathrm{\mid }{z}^n\mathrm{/}n\mathrm{\mid }=1\mathrm{/}n$, so the modulus series also diverges by [L2]; the convergence from step 2.1 is therefore conditional. No term $1\mathrm{/}0$ is formed.

For $n\ge 1$, $\mathrm{\mid }{z}^n\mathrm{/}{n}^2\mathrm{\mid }\le 1\mathrm{/}{n}^2$.

The majorant converges by [L2], so [L1] proves absolute and uniform convergence on the entire closed disc, including its boundary.

For each $n\ge 1$, ${\sup }_{z\in D}\mathrm{\mid }{z}^n\mathrm{\mid }=1$, although the supremum is not attained: real $z\uparrow 1$ makes $z^n\uparrow 1$.

If the series converged uniformly, its terms would tend uniformly to $0$, contradicting step 1.1 and the quantifiers in [L1].

Nevertheless [L2] gives pointwise convergence throughout $D$, so this is a counterexample to uniform convergence on the whole open disc.

Both coefficient sequences have root limsup $1$, so [L1] gives radius $1$ to both series.

At $z=1$, the first series has constant term sequence $1$ and diverges by [L3], while the second converges by [L2] with $p=2$.

Thus equal radii do not determine even convergence at the boundary point $1$.

Every coefficient root is $0$ or $1$, since each $c_m$ is $0$ or $1$. By [L3] every $n!$ is a nonzero natural, hence $n!\ge 1$, and the recursion clause then gives $\sigma (n)!=n!\cdot \sigma (n)\ge \sigma (n)$; so $m!\ge m$ for every $m\ge 1$ and the set $\{n!:n\ge 2\}$ of indices carrying $c_m=1$ is unbounded. The value $1$ therefore occurs at arbitrarily large indices, so the root limsup is $1$.

By [L1] the radius is $1$. At $z=1$, the terms $c_m$ do not tend to $0$ because $c_{n!}=1$ for every $n\ge 2$, so [L2] gives divergence.

The finite geometric identity with ratio $-x^2$ gives the displayed series for $\mathrm{\mid }x\mathrm{\mid }<1$; its coefficients at even indices have modulus $1$, so [L1] gives radius $1$.

The hypothesis of [L2] is that the inputs are already differentiable, so the induction needs a base: by [L5] the constant and identity functions are differentiable everywhere, and [L2] applied to sums and products makes every real polynomial differentiable everywhere, $1+x^2$ among them. Since $1+x^2>0$ for every real $x$, the quotient clause of [L2] applies at every point, and an induction on the order — each step differentiating a quotient of polynomials with denominator a positive power of $1+x^2$, which [L5] and [L2] make differentiable — expresses every derivative of $f$ as such a quotient. [L3] then makes each derivative continuous.

Therefore $f$ is smooth by [L4] while its Maclaurin series has finite radius.

Direct calculation gives $\mathrm{\mid }{z}_k{\mathrm{\mid }}^2=1-2\mathrm{/}k+5\mathrm{/}(4k^2)<1$ for $k\ge 2$, and $\mathrm{\mid }1-z_k\mathrm{\mid }=\sqrt{5}\mathrm{/}(2k)$.

Since $1-\mathrm{\mid }{z}_k\mathrm{\mid }=(1-\mathrm{\mid }{z}_k{\mathrm{\mid }}^2)\mathrm{/}(1+\mathrm{\mid }{z}_k\mathrm{\mid })$, step 1.1 gives a uniform bound on $\mathrm{\mid }1-z_k\mathrm{\mid }\mathrm{/}(1-\mathrm{\mid }{z}_k\mathrm{\mid })$, so $(z_k)$ lies in one Stolz region and tends to $1$.

Apply [L2] to the series in [L1]. The nonzero imaginary part makes the approach nonradial.

At $z=1\mathrm{/}2$, the finite geometric identity shows that the partial sums tend to $2$, so the series converges at a noncentral point.

At $z=2$, its terms $2^n$ do not tend to $0$, so the series diverges by [L2].

Hence the claim is false. Consistently, [L1] gives this series radius $1$: one interior convergence point does not force an infinite radius.

The sourced proposition [L1] directly contradicts the asserted global boundedness of both functions.

Thus the claim that both functions are bounded on $\mathbb{C}$ is false.