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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 if the original formula is valid for $|x| \le 1$ or only for $|x| <1$.

3rd Question: I had 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.

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.

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 if the original formula is valid for $|x| \le 1$ or only for $|x| <1$.

3rd Question: I had 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.

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 if the original formula is valid for $|x| \le 1$ or only for $|x| <1$.

3rd Question: I had 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.

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: 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 if the original formula is valid for $|x| \le 1$ or only for $|x| <1$.

3rd Question: I had 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.

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 if the original formula is valid for $|x| \le 1$ or only for $|x| <1$.

3rd Question: I had 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.