Can we re-write Newton's Binomial formula as a power series in $\ r\ $ without any problems? Newton's Generalised Binomial theorem states that if $\ x\ $ and $\ y\ $ are real numbers with $\ \vert x \vert > \vert y \vert\ (\text{note that } \left\vert \frac{y}{x} \right\vert < 1),\ $ and $\ r\ $ is any complex number, one has
$$ (x+y)^r = \sum_{k=0}^{\infty} \binom{r}{k} x^{r-k} y^k = x^r + \frac{r}{1!} x^{r-1} y^1 + \frac{r(r-1)}{2!} x^{r-2} y^2 + \frac{r(r-1)(r-2)}{3!} x^{r-3} y^3 + \ldots\ .$$
If we restrict $\ r\ $ to be a real number, and view this as a power series in $\ r\ $ whilst preserving the order of terms, then:

*

*The constant term is $\ x^r.$

*The linear coefficient, i.e. the coefficient of $\ r,\ $ whilst preserving the order of terms in Newton's formula, is

$$\frac{1}{1} x^{r-1} y^1 - \frac{1}{2} x^{r-2} y^2 + \frac{1}{3} x^{r-3} y^3 - \frac{1}{4} x^{r-2} y^2 + \ldots = x^r\left( \frac{1}{1!}\frac{y}{x} - \frac{1}{2!}\left( \frac{y}{x}\right)^2 + \frac{3}{3!}\left( \frac{y}{x}\right)^3 - \frac{6}{4!} \left( \frac{y}{x}\right)^4 + \ldots \right)$$
I noticed that this is equal to $\ x^r \ln \left( 1 + \frac{y}{x} \right),\ $ although I'm not sure this fact is relevant.

*

*The quadratic coefficient, i.e. the coefficient of $\ r^2,\ $ again whilst preserving the order of terms in Newton's formula, is

$$ x^r\left(  \frac{1}{1!} \left( \frac{y}{x}\right) - \frac{3}{3!} \left( \frac{y}{x}\right)^2 +\frac{11}{4!} \left( \frac{y}{x}\right)^3 - \frac{50}{6!} \left( \frac{y}{x}\right)^4 + \ldots \right)$$
The coefficients of $\ \left(\frac{y}{x}\right)^n\ $ are Stirling numbers of the first kind divided by factorials.
Now I'm sure all these series converge, by a little bit of work (which I'm working on...) and then the Alternating Series Test and/or/ Ratio Test, since $\ \left\vert \frac{y}{x} \right\vert < 1.\ $ So each coefficient of $\ r^k\ $ exists.
My main question is: is it true that
$$ (x+y)^r = x^r + x^r\left( \frac{1}{1!}\frac{y}{x} - \frac{1}{2!}\left( \frac{y}{x}\right)^2 + \frac{3}{3!}\left( \frac{y}{x}\right)^3 - \frac{6}{4!} \left( \frac{y}{x}\right)^4 + \ldots \right) r + x^r\left( \frac{1}{1!} \left( \frac{y}{x}\right) - \frac{3}{3!} \left( \frac{y}{x}\right)^2 +\frac{11}{4!} \left( \frac{y}{x}\right)^3 - \frac{50}{6!} \left( \frac{y}{x}\right)^4 + \ldots \right) r^2 + \ldots\ ?$$
This can be re-written more nicely as:
$$ \left(1+\left(\frac{y}{x}\right) \right)^r = g(r):= 1 + \left( \frac{1}{1!}\frac{y}{x} - \frac{1}{2!}\left( \frac{y}{x}\right)^2 + \frac{3}{3!}\left( \frac{y}{x}\right)^3 - \frac{6}{4!} \left( \frac{y}{x}\right)^4 + \ldots \right) r + \left(  \frac{1}{1!} \left( \frac{y}{x}\right) - \frac{3}{3!} \left( \frac{y}{x}\right)^2 +\frac{11}{4!} \left( \frac{y}{x}\right)^3 - \frac{50}{6!} \left( \frac{y}{x}\right)^4 + \ldots \right) r^2 + \ldots$$
which is clearly a power series in $\ r.$
I'm not even sure if $\ g(r)\ $ exists for all values of $\ r\ $ let alone if it is equal to $\ \left(1+\left(\frac{y}{x}\right) \right)^r.$
I'm not sure if the Riemann Series Theorem has anything to say about this, since this is technically not a simple rearrangement of the terms in Newton's formula: more specifically, we have not permuted the terms...
 A: $$ (x+y)^r = \sum_{k=0}^{\infty} \binom{r}{k} x^{r-k} y^k = x^r + \frac{r}{1!} x^{r-1} y^1 + \frac{r(r-1)}{2!} x^{r-2} y^2 + \frac{r(r-1)(r-2)}{3!} x^{r-3} y^3 + \ldots\ .$$
Dividing through by the constant $\ x^r,\ $ we see that the above series converges if and only if
$$ \left(1+\frac{y}{x}\right)^r = \sum_{k=0}^{\infty} \binom{r}{k} \left(\frac{y}{x}\right)^k = \underbrace{1}_{a_0} + \underbrace{\frac{r}{1!}}_{a_1} \left(\frac{y}{x}\right)^1 + \underbrace{\frac{r(r-1)}{2!} }_{a_2} \left(\frac{y}{x}\right)^2 + \underbrace{\frac{r(r-1)(r-2)}{3!} }_{a_3} \left(\frac{y}{x}\right)^3 + \ldots $$
converges, and we know that this Newton's Generalised Binomial Expansion converges for all real $\ r\in\mathbb{R}.$
However, this does not immediately imply the series of expanded brackets terms
$$ \underbrace{1}_{b_0} + \underbrace{\frac{r}{1!}}_{b_1}\left(\frac{y}{x}\right)^1 + \underbrace{- \frac{r}{2!} }_{b_2} \left(\frac{y}{x}\right)^2 + \underbrace{\frac{r^2}{2!} }_{b_3} \left(\frac{y}{x}\right)^2 + \underbrace{\frac{2r}{3!} }_{b_4} \left(\frac{y}{x}\right)^3 + \underbrace{- \frac{3r^2}{3!} }_{b_5} \left(\frac{y}{x}\right)^3 + \underbrace{\frac{r^3}{3!} }_{b_6} \left(\frac{y}{x}\right)^3 + \ldots$$
is equal to $\ \left(\ 1+\frac{y}{x}\right)^r:\ $ we have not yet even established if $\ \displaystyle\sum_k b_k \left(\frac{y}{x}\right)^k\ $ converges.
We proceed by showing that $\ \displaystyle\sum_k b_k \left(\frac{y}{x}\right)^k\ $ converges absolutely, as follows.
$$ \underbrace{1}_{c_0} + \underbrace{ \frac{ \vert r \vert }{1!} }_{c_1} \left\vert \frac{y}{x}\right\vert^1 + \underbrace{ \frac{ \vert r \vert ( \vert r \vert + 1 ) }{2!} }_{c_2} \left\vert\frac{y}{x}\right\vert^2 + \underbrace{ \frac{ \vert r \vert (  \vert r \vert + 1 )(  \vert r \vert + 2 ) }{3!} }_{c_3} \left\vert\frac{y}{x}\right\vert^3 + \ldots $$
is a series of positive terms; this series converges by the ratio test, because
$$ \left \vert \frac{c_{k+1}\left\vert\frac{y}{x}\right\vert^{k+1} }{c_{k} \left\vert\frac{y}{x}\right\vert^k} \right \vert = \left\vert \frac{\left \vert r \right \vert + k}{k+1} \right\vert \left\vert \frac{y}{x} \right\vert \overset{k \to \infty}{\to} \vert 1 \vert \left\vert \frac{y}{x} \right\vert = \left\vert \frac{y}{x} \right\vert < 1.  $$
So, splitting $\ c_k\ $ in the previous series up into individual terms by expanding brackets, we get:
$$ 1 + \frac{ \vert r \vert }{1!} \left\vert \frac{y}{x}\right\vert^1 + \frac{ \vert r \vert }{2!} \left\vert \frac{y}{x}\right\vert^2 + \frac{ \vert r \vert ^2 }{2!} \left\vert \frac{y}{x}\right\vert^2 + \frac{ 2 \vert r \vert }{3!} \left\vert\frac{y}{x}\right\vert^3 + \frac{ 3 \vert r \vert ^2 }{3!} \left\vert\frac{y}{x}\right\vert^3 + \frac{ \vert r \vert ^3 }{3!} \left\vert\frac{y}{x}\right\vert^3 + \ldots,\qquad (1) $$
which is a series of positive terms whose limit of partial sums must be equal to $\ \displaystyle\sum_{k=0}^{\infty}\ c_{k} \left\vert\frac{y}{x}\right\vert^k,\ $ by the Monotone Convergence Theorem. We can allow any of the terms in $\ (1)\ $ to be negative and it will still converge. In other words, $\ \displaystyle\sum_{k=0}^{\infty} b_k \left( \frac{y}{x}\right)^k\ $ converges absolutely.
The first consequence of this is that $\ \displaystyle\sum_{k=0}^{\infty} b_k \left( \frac{y}{x}\right)^k\ = \sum_{k=0}^{\infty} \binom{r}{k} \left(\frac{y}{x}\right)^k,\ $ for all values of $\ r\in\mathbb{R},\ $ because simply grouping together terms in the series $\ \displaystyle\sum_{k=0}^{\infty} b_k \left( \frac{y}{x}\right)^k\ $ without changing the order of terms, doesn't change the fact that it converges, nor the value it converges to.
Secondly, by Fubini's theorem for infinite series (and not Riemann Series Theorem), the fact that $\ \displaystyle\sum_{k=0}^{\infty} b_k \left( \frac{y}{x}\right)^k\ $ converges absolutely means that $\ \displaystyle\sum_{k=0}^{\infty} b_k \left( \frac{y}{x}\right)^k\ $ is indeed equal to
$$ g(r):= 1 + \left( \frac{1}{1!}\frac{y}{x} - \frac{1}{2!}\left( \frac{y}{x}\right)^2 + \frac{3}{3!}\left( \frac{y}{x}\right)^3 - \frac{6}{4!} \left( \frac{y}{x}\right)^4 + \ldots \right) r + \left(  \frac{1}{1!} \left( \frac{y}{x}\right) - \frac{3}{3!} \left( \frac{y}{x}\right)^2 +\frac{11}{4!} \left( \frac{y}{x}\right)^3 - \frac{50}{6!} \left( \frac{y}{x}\right)^4 + \ldots \right) r^2 + \ldots\ $$
for all values of $\ r\in\mathbb{R}.$
Thus we finally have our result:
$$\ (x+y)^r = \sum_{k=0}^{\infty} \binom{r}{k} x^{r-k} y^k = x^r + \frac{r}{1!} x^{r-1} y^1 + \frac{r(r-1)}{2!} x^{r-2} y^2 + \frac{r(r-1)(r-2)}{3!} x^{r-3} y^3 + \ldots\ = \sum_{k=0}^{\infty} b_k \left( \frac{y}{x}\right)^k\ = g(r):= 1 + \left( \frac{1}{1!}\frac{y}{x} - \frac{1}{2!}\left( \frac{y}{x}\right)^2 + \frac{3}{3!}\left( \frac{y}{x}\right)^3 - \frac{6}{4!} \left( \frac{y}{x}\right)^4 + \ldots \right) r + \left(  \frac{1}{1!} \left( \frac{y}{x}\right) - \frac{3}{3!} \left( \frac{y}{x}\right)^2 +\frac{11}{4!} \left( \frac{y}{x}\right)^3 - \frac{50}{6!} \left( \frac{y}{x}\right)^4 + \ldots \right) r^2 + \ldots $$
for all values of $\ r\in\mathbb{R}.$
