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

1. Apr 4, 2013

### Shoelace Thm.

1. The problem statement, all variables and given/known data
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 $\sum_{k=0}^{n} \binom{n+1}{k} B_k = 0$.

2. Relevant equations

3. 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.

2. Apr 4, 2013

### Shoelace Thm.

Does anyone have any suggestions?

3. Apr 4, 2013

l'Hôpital?

4. Apr 4, 2013

### Staff: Mentor

You can get the first part pretty easily by long division of x by ex - 1 (= x + x2/2! + x3/3! + ... + xn/n! + ...).

5. Apr 4, 2013

### Shoelace Thm.

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.

6. Apr 4, 2013

### Staff: Mentor

Sure it is.
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.
The coefficient of the nth term is Bn, pretty much by definition. What you need to prove is that $$\sum_{k=0}^{n} \binom{n+1}{k} B_k = 0$$
Well, maybe, but what I've suggested is what I would try.

7. Apr 4, 2013

### vela

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

8. Apr 5, 2013

### Shoelace Thm.

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.

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).

9. Apr 5, 2013

### vela

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

10. Apr 5, 2013

### Shoelace Thm.

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

11. Apr 5, 2013

### vela

Staff Emeritus
$$\frac{x}{e^x-1} = \frac{1}{\frac{e^x-1}{x}}.$$ 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.

12. Apr 6, 2013

### Shoelace Thm.

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.

13. Apr 7, 2013

### Shoelace Thm.

Any thoughts?