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\newtheorem{theorem}{Theorem}

\theoremstyle{remark}
\newtheorem*{remark}{Remark}

\begin{document}

\title{Abel's theorem for series of powers of functions}

\author{David Milovich}

\maketitle

\begin{theorem}
Let $g(x)$ be a real function continuous on a nonempty open interval $(a,b)$ (and possibly continuous elsewhere);
let $c_0,c_1,c_2,\ldots$ be a sequence of constants.
Suppose that $\liml_{k\rightarrow\infty}\left\lvert\frac{c_{k+1}g(x)^{k+1}}{c_kg(x)^k}\right\rvert$ 
exists and is $<1$ for all $x$ in $(a,b)$, except where $g(x)=0$.

If $f(x)$ and $g(x)$ are continuous from the left at $b$ and
$f(x)=\suml_{k=0}^\infty c_k(g(x))^k$ for all $x$ in $(a,b)$,
then $f(b)=\suml_{k=0}^\infty c_k(g(b))^k$.

Likewise, if $f(x)$ and $g(x)$ are continuous from the right at $a$ and
$f(x)=\suml_{k=0}^\infty c_k(g(x))^k$ for all $x$ in $(a,b)$,
then $f(a)=\suml_{k=0}^\infty c_k(g(a))^k$.
\end{theorem}
\begin{proof}
By symmetry, we only need to prove the first half of the theorem;
the second half follows the first half applied 
to $p(x)=g(a+b-x)$ in place of $g(x)$.

Let $S(x)=\suml_{k=0}^\infty c_k(g(x))^k$.
If $g(x)=0$ on all of $(a,b)$, then $f(x)=c_0$ on all of $(a,b)$,
and, by continuity, $g(b)=0$ and $f(b)=c_0$, so
$f(b)=c_0=S(b)$.  Therefore, 
we may assume $g(x)\not=0$ for some $x$ in $(a,b)$.
Hence, at such an $x$, $1>
\liml_{k\rightarrow\infty}\left\lvert\frac{c_{k+1}g(x)^{k+1}}{c_kg(x)^k}\right\rvert
=\lvert g(x)\rvert\liml_{k\rightarrow\infty}\left\lvert\frac{c_{k+1}}{c_k}\right\rvert$;
let $L=\liml_{k\rightarrow\infty}\left\lvert\frac{c_{k+1}}{c_k}\right\rvert$.

Choose $q$ in $(a,b)$.
If $L=0$, then the ratio-test limit for $S(x)$ will be $0$,
or $g(x)=0$, for all $x$ in $[q,b]$, so $S(x)$ will be
continuous on $[q,b]$, so $S(b)=\lim_{x\rightarrow b^-}S(x)=\lim_{x\rightarrow b^-}f(x)=f(b)$.
Therefore, we may assume $L>0$.
Similarly, if $\lvert g(b)\rvert<1/L$, then the ratio-test limit for $S(x)$ will be $<1$,
or $g(x)=0$, for all $x$ in $[q,b]$, so again $S(b)=f(b)$.
Therefore, we may assume $g(b)\geq 1/L$.
%In particular, $g(b)\not=0$.

If $\lvert g(b)\rvert>1/L$, then, by the Intermediate Value Theorem,
if we choose $r$ in $(1/L,\lvert g(b)\rvert)$, then $\lvert g(y)\rvert=r$ for some $y$ in $(q,b)$,
for $\lvert g\rvert$ is continuous on $[q,b]$ and $\lvert g(q)\rvert<1/L<r<\lvert g(b)\rvert$.
However, in this case $g(y)\not=0$ and $L\lvert g(y)\rvert>1$, 
in contradiction with the ratio-test limit for $S(x)$ being $<1$ 
for all $x$ in $(a,b)$ where $g(x)\not=0$.
Therefore, $\lvert g(b)\rvert=1/L$.

With the easier cases dispensed with, 
our strategy will now be to prove that $S(b)$ is infinitely close to
$S(x)$ for some $x$ in $(a,b)$ that is infinitely close to $b$.
Since $S(x)=f(x)$ and $f$ is continuous from the left at $b$, 
$x\approx b$ will imply that $S(b)\approx f(x)\approx f(b)$.
Since $S(b)$ and $f(b)$ are both reals,
they can only be infinitely close if they are equal,
which is exactly what we wish to prove.
Therefore, we just need to find $x$ such
that $b\approx x<b$ and $S(x)\approx S(b)$.

Set $d_k=c_k(g(b))^k$ and $h(x)=g(x)/g(b)$, so that $S(x)=\suml_{k=0}^\infty d_k(h(x))^k$,
$h(b)=1$, and $S(b)=\suml_{k=0}^\infty d_k$.
Since $h(b)=1$ and $h(x)=g(x)/g(b)$ is continuous on $(a,b]$,
if $J$ is an infinite positive hyperinteger,
then $h(b-1/J)\approx h(b)=1$, so $h(b-1/J)>0$.
Fixing such a $J$, let $I$ be the greatest hyperinteger
less than $J$ for which there is $h(b-z)\leq 0$
for some positive $z<1/I$;
if no such hyperinteger exists, set $I=1$.  
Since $h(b-z)\approx h(b)=1$ for all positive $z\approx 0$,
$I$ cannot be infinite, so $I$ is finite and, 
setting $t=b-1/I$, $h(x)>0$ for all $x$ in $(t,b]$.
We will use this fact later.

Set $s_k=d_0+d_1+\cdots+d_{k-1}+d_k$, so that $s_k-s_{k-1}=d_k$ and $S(b)=\lim_{n\rightarrow\infty}s_n$.
Therefore, 
\begin{align*}
\suml_{k=0}^n d_k h^k
&=d_0+d_1h+d_2h^2+d_3h^3+\cdots+d_{n-1}h^{n-1}+d_nh^n\\
&=s_0+(s_1-s_0)h+(s_2-s_1)h^2+(s_3-s_2)h^3+\cdots+(s_{n-1}-s_{n-2})h^{n-1}+(s_n-s_{n-1})h^n\\
&=s_0(1-h)+s_1(h-h^2)+s_2(h^2-h^3)+\cdots+s_{n-1}(h^{n-1}-h^n)+s_nh^n\\
&=(1-h)(s_0+s_1h+s_2h^2+\cdots+s_{n-1}h^{n-1})+s_nh^n\\
&=(1-h)\suml_{k=0}^{n-1}s_k h^k+s_nh^n
\end{align*}
For all $x$ in $(a,b)$, $\lvert h(x)\rvert=\left\lvert\frac{g(x)}{g(b)}\right\rvert=\lvert g(x)L\rvert<1$,
so $\lim_{n\rightarrow\infty}h^n(x)=0$, so
\begin{align*}
S(x)=\liml_{n\rightarrow\infty}\suml_{k=0}^n d_kh^k(x)
&=\liml_{n\rightarrow\infty}\left((1-h(x))\suml_{k=0}^{n-1}s_k h^k(x)+s_nh^n(x)\right)\\
&=(1-h(x))\suml_{k=0}^{\infty}s_k h^k(x)+S(b)\cdot 0
\end{align*}
Therefore, again using the fact that $\lvert h(x)\rvert<1$,
\begin{align*}
S(x)-S(b)
&=(1-h(x))\suml_{k=0}^{\infty}s_k h^k(x)-S(b)\\
&=(1-h(x))\left(\suml_{k=0}^{\infty}s_k h^k(x)-\frac{S(b)}{1-h(x)}\right)\\
&=(1-h(x))\left(\suml_{k=0}^{\infty}s_k h^k(x)-S(b)\suml_{k=0}^{\infty}h^k(x)\right)\\
&=(1-h(x))\suml_{k=0}^{\infty}(s_k-S(b))h^k(x)\\
&=(1-h(x))\suml_{k=0}^{M-1}(s_k-S(b))h^k(x)+(1-h(x))\suml_{k=M}^{\infty}(s_k-S(b))h^k(x)
\end{align*}
for all positive $M$.


Choose $M$ to be infinitely large.  Let $N$ be the greatest hyperinteger 
(if it exists) for which $\lvert s_k-S(b)\rvert<\frac{1}{N}$ for all $k\geq M$.  
Since $\liml_{n\rightarrow\infty}s_n=S(b)$, 
$s_k\approx S(b)$ for all infinite $k$, so $s_k\approx S(b)$ for all $k\geq M$,
so $N$ exists and must be infinite.
Therefore, for all $x$ in $(t,b)$ (where $0<h(x)<1$),
\begin{align*}
\left\lvert(1-h(x))\suml_{k=M}^{\infty}(s_k-S(b))h^k(x)\right\rvert
&<\left\lvert\frac{1-h(x)}{N}\right\rvert\suml_{k=M}^\infty\lvert h^k(x)\rvert\\
&=\frac{1-h(x)}{N}(h^M(x)+h^{M+1}(x)+h^{M+2}(x)+\cdots)\\
&=\frac{(1-h(x))h^M(x)}{N}(1+h(x)+h^{2}(x)+\cdots)\\
&=\frac{(1-h(x))h^M(x)}{N}\frac{1}{1-h(x)}\\
&=\frac{h^M(x)}{N}.
\end{align*}
Since $\lvert h(x)\rvert<1$, $\lvert h^M(x)\rvert < 1$, so
$\left\lvert(1-h(x))\suml_{k=M}^{\infty}(s_k-S(b))h^k(x)\right\rvert<\frac{1}{N}$.
Therefore, $$(1-h(x))\suml_{k=M}^{\infty}(s_k-S(b))h^k(x)\approx 0.$$
Hence, $S(x)-S(b)\approx(1-h(x))\suml_{k=0}^{M-1}(s_k-S(b))h^k(x)+0$.
%for all $x$ in $(t,b)$.

Let $P$ be the maximum value appearing in the (hyper)list 
$\lvert s_0-S(b)\rvert, \lvert s_1-S(b)\rvert,\ldots,\lvert s_{M-1}-S(b)\rvert$.
$\lvert s_i-S(b)\rvert$ is finite for all finite $i$,
and $s_i\approx S(b)$ for all infinite $i$, 
so $\lvert s_i-S(b)\rvert$ is finite for all $i$.
Therefore, $P$ is finite, so we can use $P$ 
to establish a useful upper bound on $\lvert S(x)-S(b)\rvert$:
\begin{align*}
\lvert S(x)-S(b)\rvert
&\approx\left\lvert(1-h(x))\suml_{k=0}^{M-1}(s_k-S(b))h^k(x)\right\rvert\\
&\leq(1-h(x))P\sum_{k=0}^{M-1}h^k(x)\\
&=(1-h(x))P\frac{1-h^M(x)}{1-h(x)}\\
&=P(1-h^M(x)).
\end{align*}

Our goal is to show that $S(x)\approx S(b)$ for some $x$ with $b\approx x<b$,
so if we can show that $h^M(x)\approx 1$ for some such $x$, 
then we will have $\lvert S(x)-S(b)\rvert\approx P(1-h^M(x))\approx 0$,
which will imply $S(x)\approx S(b)$.
When $b\approx x<b$, $h(x)\approx h(b)=1$,
so $h^m(x)\approx 1$ for all real $m$. Therefore,
for every positive real $\delta$ and real $m$,
there is a hyperreal $x$ such that
$b-\delta<x<b$ and $\lvert 1-h^m(x)\rvert<\delta$.
By Transfer, for every positve real $\delta$ and real $m$, 
there is a real $x$ such that
$b-\delta<x<b$ and $\lvert 1-h^m(x)\rvert<\delta$.
By Transfer again, 
for every positive hyperreal $\delta$ and hyperreal $m$,
there is a hyperreal $x$ such that
$b-\delta<x<b$ and $\lvert 1-h^m(x)\rvert<\delta$.
In particular, if we choose $\delta$ to be positive infinitesimal,
then there is a hyperreal $x$ such that
$b-\delta<x<b$ and $\lvert 1-h^M(x)\rvert<\delta$.
Thus, there is a hyperreal $x$ such that
$b\approx x<b$ and $h^M(x)\approx 1$, so the proof is complete.
\end{proof}

\begin{remark}
The above proof can be extended to show that 
$\liml_{b\rightarrow 1^-}S(x)=S(b)$,
even if $f$ is not continuous from the left at $b$.  
To do this, we just need to show that $S(x)\approx S(b)$
for {\it all} $x$ satisfying $b\approx x<b$.
The above proof showed that if $M$ is a positive infinite hyperinteger,
then $h^M(x)\approx 1$ implies $S(x)\approx S(b)$
for all x satisfying $t<x<b$.
Therefore, it is enough to show that 
for all $x$ satisfying $b\approx x<b$,
we can find an infinite positive hyperinteger $M$ 
such that $h^M(x)\approx 1$.

By the definition of derivative, if $0\approx\eps\not=0$, then 
$\frac{\ln(1+\eps)}{\eps}\approx\ln'(1)=\frac{1}{1}=1$.
So, given any $x$ satisfying $b\approx x<b$,
we have $1\approx h(x)<1$, so $0\approx h(x)-1<0$, so we can 
set $\eps=h(x)-1$ to get $\frac{\ln(1+(h(x)-1))}{h(x)-1}\approx 1$,
which implies $-1\approx\frac{1}{1-h(x)}\ln(h(x))$,
which in turn implies 
$$\ee^{-1}\approx\ee^{(\ln(h(x)))(1/(1-h(x)))}=\left(\ee^{\ln(h(x))}\right)^{1/(1-h(x))}=h(x)^{1/(1-h(x))}.$$
Since $1\approx h(x)<1$ implies $0\approx 1-h(x)>0$,
$1/\sqrt{1-h(x)}$ is positive infinite;
let $M$ be the greatest hyperinteger $\leq 1/\sqrt{1-h(x)}$.
Therefore, since $0<h(x)<1$, we have
$$1>h(x)^M\geq h(x)^{1/\sqrt{1-h(x)}}
=h(x)^{\sqrt{1-h(x)}/(1-h(x))}
=\left(h(x)^{1/(1-h(x))}\right)^{\sqrt{1-h(x)}}
\approx(\ee^{-1})^0=1,$$
so indeed $h^M(x)\approx 1$.
\end{remark}

\end{document}
