Binomial Transform and Ordinary Generating Functions

Question

I had a couple of simple doubts about the following Theorem from Wikipedia's Binomial Transform page, which I have not been able to solve by searching for information over the Internet:

Let

$$f(x)=\sum_{n=0}^\infty a_n x^n$$

And

$$g(x)=\sum_{n=0}^\infty s_n x^n$$

Where $s_n$ is the binomial transform of $a_n$. Then,

$$g(x)=\frac{1}{1-x} f \left ( \frac{-x}{1-x} \right )$$

1st Question (already solved): After some time looking for information about it on the Internet, I have not found any proof for this Theorem, and I do not have access to all of Wikipedia's suggested bibliography. Where can I find this proof?

2nd Question: Does that formula still hold for:

$$g(1)=\lim_{x \to 1^-} \frac{1}{1-x} f \left ( \frac{-x}{1-x} \right )$$

As we are working with Ordinary Generating Functions, I do not know wether the original formula is valid for $|x| \le 1$ or only for $|x| <1$.

3rd Question: I am having some trouble when working with this specific series:

Let

$$a_n= \frac{B_{n+1}}{8^n (n+1)!}$$

So that

$$s_n= \sum_{k=0}^n (-1)^k {n \choose k} \frac{B_{k+1}}{8^k (k+1)!}$$

And let both $h(x)$ and $i(x)$ be defined as $f(x)$ and $g(x)$ but for these specific sequences above. Then,

$$h(x)=\sum_{n=0}^\infty \frac{B_{n+1}}{ (n+1)!}{\left ( \frac{x}{8} \right )}^n = \frac{8}{x} \sum_{n=1}^\infty \frac{B_{n}}{ (n)!}{\left ( \frac{x}{8} \right )}^n = \frac{8}{x} \left ( \sum_{n=0}^\infty \frac{B_{n}}{ (n)!}{\left ( \frac{x}{8} \right )}^n -1 \right )$$

By the generating function of Bernoulli Numbers (and taking $B_1=\frac{1}{2}$),

$$ h(x)= \frac{8}{x} \left ( \frac{\frac{x}{8}}{1-e^{-\frac{x}{8}}}-1 \right ) =\frac{1}{1-e^{-\frac{x}{8}}} - \frac{8}{x}$$

Then, as

$$i(x)=\frac{1}{1-x} f \left ( \frac{-x}{1-x} \right )$$

We have that:

$$i(x)=\frac{1}{1-x} \left ( \frac{1}{1-e^{-\frac{\frac{-x}{1-x}}{8}}} - \frac{8}{\frac{-x}{1-x}} \right )$$

$$i(x)= \frac{1}{(1-x)(1-e^{\frac{x}{8(1-x)}})} + \frac{8}{x}$$

However, the original series for $i(1)$ diverges but $$\lim_{x \to 1^-} \frac{1}{(1-x)(1-e^{\frac{x}{8(1-x)}})} + \frac{8}{x} = 8$$

I also have some trouble when choosing some values for $x$ close to $1$.

Assuming that the answer to the 2nd question is yes, is there any flaw in my work? Is there any concept that I am missunderstanding?

Thank you.

Edit: to clarify a bit what I want here. Intuition tells me that, for some $x<1$ very close to $1$, the formula above should be valid (while on the other hand, some small computations give numerical evidence that it isn't). Moreover, the following limit should exist:

$$\lim_{x \to 1^-} i(x)-\frac{1}{1-x} h \left ( \frac{-x}{1-x} \right ) = 0 $$

While, substituting $\frac{1}{1-x} h \left ( \frac{-x}{1-x} \right )$ with $\frac{1}{(1-x)(1-e^{\frac{x}{8(1-x)}})} + \frac{8}{x}$ as obtained before, we gent that the limit does not exist, since $i(1)$ diverges and the subtrahend equals 8.

Answer 1

The formula is $g(x) = f(-x/(1-x))/(1-x)$ in one convention. To answer your 1st question, the proof essentially depends on applying the binomial expansion to $f(x)$ using the formula $$\left(\frac{1}{1-x}\right)\left(\frac{-x}{1-x}\right)^n = (-x)^n\left(1 + {n +1 \choose 1}x+{n+2 \choose 2}x^2+\dots\right). $$ To answer your 2nd question, the default case of generating functions is that they are "formal" power series and need not converge anywhere expect at $x=0$, so you can't draw any conclusions about $g(1)$ without further investigation.

For your 3rd question, you encountered a case where $g(1)$ as series does not converge, and the $g(x)$ expression goes to infinity as $x\to 1^-$. In order to put this in perspective, consider $f(x)=x$ where $g(x)=-x/(1-x)^2$ so $g(x)\to -\infty$ as $x\to 1$ and the series does not converge.

Answer 2

With respect to OPs first question:

The transformation is known as Euler transform of a series \begin{align*} A(t)=\sum_{n= 0}^\infty a_nt^n\qquad\qquad \frac{1}{1-t}A\left(\frac{-t}{1-t}\right)=\sum_{n= 0}^\infty \left(\sum_{j=0}^n\binom{n}{j}(-1)^ja_j\right)t^n \end{align*} This transformation formula together with a proof can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev. See remark 2 with $\lambda=1, \mu=-1$.

Hint: This answer might be useful.

Answer 3

Speaking in terms of Formal Series (as already noted in another answer) it is easy, starting from the known Power series $$ {{z^n } \over {\left( {1 - z} \right)^{n + 1} }} = \sum\limits_{0\, \le \,k} {\left( \matrix{ k \cr n \cr} \right)} \;z^k \quad {\rm integer n} \ge {\rm 0 } $$ to demonstrate that $$ \eqalign{ & \sum\limits_{0\, \le \,n} {a_{\,n} {{z^n } \over {\left( {1 - z} \right)^{n + 1} }}} = {1 \over {\left( {1 - z} \right)}}\sum\limits_{0\, \le \,n} {a_{\,n} \left( {{z \over {1 - z}}} \right)^n } = {1 \over {\left( {1 - z} \right)}}f\left( {{z \over {1 - z}}} \right) = \cr & = \sum\limits_{0\, \le \,n} {a_{\,n} \sum\limits_{0\, \le \,k} {\left( \matrix{ k \cr n \cr} \right)} \;z^k } = \sum\limits_{0\, \le \,k} {\left( {\sum\limits_{0\, \le \,n} {\left( \matrix{ k \cr n \cr} \right)a_{\,n} } } \right)\;z^k } = \cr & = \sum\limits_{0\, \le \,k} {s_{\,k} \;z^k } = g(z) \cr} $$

and further variations by introducing $(-1)^n$ and $(-1)^{n-k}$