# How to expand the series $\dfrac{1}{e^x-1}$

How to expand the series $$\dfrac{1}{e^x-1}$$

I am looking for a way to expand this series $$\dfrac{1}{e^x-1}$$. The goal is to be able to finally derive the expansion for $$\dfrac{x}{e^x-1}$$ by multiplying $$x$$ with the numerator of the first series.

So far, I am stuck. I try to do this as followed:

$$1+\dfrac{(-1)(-e^x)}{1!}+\dfrac{(-1)(-2)(-e^x)^2}{}+\dfrac{(-1)(-2)(-3)(-e^x)^3}{3!}...$$

This form doesn't look right, Wolfram indicates that the power series for

$$\dfrac{1}{e^x-1}=\dfrac{1}{x}-\dfrac{1}{2}+\dfrac{x}{12}-\dfrac{x^3}{720}...$$

$$\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{x^2}{12}-\dfrac{x^4}{720}...$$

I think my approach to get the series is right, but I don't know how to make $$\dfrac{1}{e^x-1}$$ equals the series given by Wolfram.

Can you show me a way to expand these series?

Use

$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

So that

$$\dfrac{1}{e^x-1} = \dfrac{1}{\sum_{n=0}^{\infty} \frac{x^n}{n!}-1} = \dfrac{1}{\sum_{n=1}^{\infty} \frac{x^n}{n!}}.$$

Then, what you're looking for is the (multiplicative) inverse of

$${\sum_{n=1}^{\infty} \frac{x^n}{n!}} = x + \frac{x^2}{2} + \frac{x^3}{3!} + ...$$

That is, some expression

$$\sum_{m= - \infty}^{\infty} \alpha_m x^m$$

such that

$$(\sum_{m=-\infty}^{\infty} \alpha_m x^m)({\sum_{n=1}^{\infty} \frac{x^n}{n!}}) = 1$$

In particular, for integers $$k$$ with $$k \neq 0$$ we have

$$\sum_{m+n = k} \frac{\alpha_m}{n!} = 0,$$

and also

$$\sum_{m+n = 0} \frac{\alpha_m}{n!} = 1.$$

This firstly forces that $$\alpha_m = 0$$ for $$m<{-1}$$. Is this enough of a hint for you to do the rest?

• From this part below I don't understand: $$\sum_{n= - \infty}^{\infty} \alpha_n x^n$$ – James Warthington Nov 4 '19 at 2:36
• Perhaps you could elaborate on which bit you aren't understanding? – Matt Nov 4 '19 at 2:43
• I have altered the indices. Please let me know if that makes anything clearer – Matt Nov 4 '19 at 2:44
• Why does it go from negative infinity to positive infinity? – James Warthington Nov 4 '19 at 2:44
• Because there may be negative powers of $x$, like $x^{-1}$, $x^{-2}$, and so on. – Matt Nov 4 '19 at 2:48

If $$\dfrac 1{e^x-1}=\dfrac{a_{-1}}x+a_0+a_1x+a_2x^2+a_3x^3+...$$

then $$\left(\dfrac{a_{-1}}x+a_0+a_1x+a_2x^2+a_3x^3+...\right)\left(x+\dfrac{x^2} 2+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}...\right)=1$$

so $$a_{-1}+\left(a_0+\dfrac {a_{-1}}2\right)x+\left(a_1+\dfrac{a_0}2+\dfrac{a_{-1}}{3!}\right)x^2+\left(a_2+\dfrac{a_1}2+\dfrac{a_0}{3!}+\dfrac{a_{-1}}{4!}\right)x^3$$

$$+\left(a_3+\dfrac{a_2}2+\dfrac{a_1}{3!}+\dfrac{a_0}{4!}+\dfrac{a_{-1}}{5!}\right)x^4+...=1$$

so $$a_{-1}=1$$ and $$a_0+\dfrac {a_{-1}}2=a_1+\dfrac{a_0}2+\dfrac{a_{-1}}{3!}=a_2+\dfrac{a_1}2+\dfrac{a_0}{3!}+\dfrac{a_{-1}}{4!}$$

$$=a_3+\dfrac{a_2}2+\dfrac{a_1}{3!}+\dfrac{a_0}{4!}+\dfrac{a_{-1}}{5!}=...=0.$$

Thus $$a_{-1}=1$$, so from $$a_0+\dfrac {a_{-1}}2=0$$ we get $$a_0=-\dfrac12$$,

so from $$a_1+\dfrac{a_0}2+\dfrac{a_{-1}}{3!}=0$$ we get $$a_1=\dfrac1{12}$$,

so from $$a_2+\dfrac{a_1}2+\dfrac{a_0}{3!}+\dfrac{a_{-1}}{4!}=0$$ we get $$a_2=0$$,

so from $$a_3+\dfrac{a_2}2+\dfrac{a_1}{3!}+\dfrac{a_0}{4!}+\dfrac{a_{-1}}{5!}=0$$ we get $$a_3=-\dfrac1{720},...$$

• Remark for future readers: I've gone ahead below and proven that we have no terms below $x^{-1}$ in the series expansion of the inverse, but I suppose it's obvious enough when written out. – Matt Nov 4 '19 at 2:35
• @J.W.Tanner I am lost in this part: $a_{-1}+\left(a_0+\dfrac {a_{-1}}2\right)x+\left(a_1+\dfrac{a_0}2+\dfrac{a_{-1}}{3!}\right)x^2+\left(a_2+\dfrac{a_1}2+\dfrac{a_0}{3!}+\dfrac{a_{-1}}{4!}\right)x^3$ – James Warthington Nov 4 '19 at 2:39
• @J.W.Tanner What are you doing with the coefficients there? – James Warthington Nov 4 '19 at 2:39
• are you lost getting to that line or the line after it? – J. W. Tanner Nov 4 '19 at 2:41
• @JamesWarthington It is mildly apparent to at least myself from your rather prompt replies that you aren't making much effort to work on this yourself. I suggest working for a while, and coming back with a more coherent question if you're still stuck, as opposed to "I don't understand how you go from line $x$ to line $y$. – Matt Nov 4 '19 at 3:30