Taylor series of e^x/(x-1)

• Shoelace Thm.
In summary, the problem asks for the coefficients of series that are obtained by dividing a polynomial by another series. The problem is stated in a somewhat difficult way, so it may be helpful to think of a polynomial as a series where all but a finite number of terms vanish. To solve for the coefficients of the series obtained in this fashion, one needs to know that the series is given by its Taylor series in a neighborhood of zero.f

Homework Statement

Let $g(x) = \frac{x}{e^x - 1} = \sum_{n=0}^{\infty} \frac{B_n}{n!} x^n$ be the taylor series for g about 0. Show B_0 = 1 and $\sum_{k=0}^{n} \binom{n+1}{k} B_k = 0$.

The Attempt at a Solution

$g(x) = \sum_{n=0}^{\infty} \frac{g^{(n)}(0)}{n!} x^n$, but $g^{(n)}(0)$ is always undefined at 0. So I don't see how any of these relations can hold.

Does anyone have any suggestions?

l'Hôpital?

Homework Statement

Let $g(x) = \frac{x}{e^x - 1} = \sum_{n=0}^{\infty} \frac{B_n}{n!} x^n$ be the taylor series for g about 0. Show B_0 = 1 and $\sum_{k=0}^{n} \binom{n+1}{k} B_k = 0$.

The Attempt at a Solution

$g(x) = \sum_{n=0}^{\infty} \frac{g^{(n)}(0)}{n!} x^n$, but $g^{(n)}(0)$ is always undefined at 0. So I don't see how any of these relations can hold.
You can get the first part pretty easily by long division of x by ex - 1 (= x + x2/2! + x3/3! + ... + xn/n! + ...).

That's true, but that's not very rigorous as it stands. What allows you to obtain series by dividing a polynomial by another series? Furthermore, how can you prove that the nth coefficient of the series obtained in this fashion is B_n for all n? This seems difficult to accomplish.

That's true, but that's not very rigorous as it stands.
Sure it is.
What allows you to obtain series by dividing a polynomial by another series?
There's not just one way to obtain a series. For example, the Maclaurin series for 1/(1 - x) is frequently obtained by long division.
Furthermore, how can you prove that the nth coefficient of the series obtained in this fashion is B_n for all n?
The coefficient of the nth term is Bn, pretty much by definition. What you need to prove is that $$\sum_{k=0}^{n} \binom{n+1}{k} B_k = 0$$
This seems difficult to accomplish.
Well, maybe, but what I've suggested is what I would try.

That's true, but that's not very rigorous as it stands. What allows you to obtain series by dividing a polynomial by another series? Furthermore, how can you prove that the nth coefficient of the series obtained in this fashion is B_n for all n? This seems difficult to accomplish.
Do you know about absolute convergence and its implication for the multiplication and division of series?

Do you know about absolute convergence and its implication for the multiplication and division of series?

Yes, but here it is not a series being divided by a polynomial, but the other way around. I have not really encountered this before, so I suppose the complete answer to your question is no.

The coefficient of the nth term is Bn, pretty much by definition. What you need to prove is that $$\sum_{k=0}^{n} \binom{n+1}{k} B_k = 0$$

I really don't see a good way of doing this. What we are obtaining through this division is some series for which a general term for B_n is not known (and I don't think there is some simple formula, after writing out the first few terms).

You can think of a polynomial as a series where all but a finite number of terms vanish.

Ok; do you have any suggestions for proving the second query?

Start with
$$\frac{x}{e^x-1} = \frac{1}{\frac{e^x-1}{x}}.$$ Then convert the problem into one of multiplying two series. It works out pretty easily. You don't need to solve for the ##B_k##'s explicitly.

Alright its simple enough. On a side note (although it is not necessary because it is given in the problem statement), if we were not given that g was given by its taylor series in a neighborhood of zero, how to prove that it is? The series manipulations require us knowing g is given by its Taylor series.

Any thoughts?