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:


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



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?

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

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?

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  • $\begingroup$ From this part below I don't understand: $$\sum_{n= - \infty}^{\infty} \alpha_n x^n$$ $\endgroup$ – James Warthington Nov 4 '19 at 2:36
  • $\begingroup$ Perhaps you could elaborate on which bit you aren't understanding? $\endgroup$ – Matt Nov 4 '19 at 2:43
  • $\begingroup$ I have altered the indices. Please let me know if that makes anything clearer $\endgroup$ – Matt Nov 4 '19 at 2:44
  • $\begingroup$ Why does it go from negative infinity to positive infinity? $\endgroup$ – James Warthington Nov 4 '19 at 2:44
  • $\begingroup$ Because there may be negative powers of $x$, like $x^{-1}$, $x^{-2}$, and so on. $\endgroup$ – 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 $a_{-1}+\left(a_0+\dfrac {a_{-1}}2\right)x+\left(a_1+\dfrac{a_0}2+\dfrac{a_{-1}}{3!}\right)x^2+\left(a_2+\dfrac{a_1}2+\dfrac{a_0}{3!}+\dfrac{a_{-1}}{4!}\right)x^3$


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!}$


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 $a_3=-\dfrac1{720},...$

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  • $\begingroup$ 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. $\endgroup$ – Matt Nov 4 '19 at 2:35
  • $\begingroup$ @J.W.Tanner I am lost in this part: $a_{-1}+\left(a_0+\dfrac {a_{-1}}2\right)x+\left(a_1+\dfrac{a_0}2+\dfrac{a_{-1}}{3!}\right)x^2+\left(a_2+\dfrac{a_1}2+\dfrac{a_0}{3!}+\dfrac{a_{-1}}{4!}\right)x^3$ $\endgroup$ – James Warthington Nov 4 '19 at 2:39
  • $\begingroup$ @J.W.Tanner What are you doing with the coefficients there? $\endgroup$ – James Warthington Nov 4 '19 at 2:39
  • $\begingroup$ are you lost getting to that line or the line after it? $\endgroup$ – J. W. Tanner Nov 4 '19 at 2:41
  • 2
    $\begingroup$ @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 $y$. $\endgroup$ – Matt Nov 4 '19 at 3:30

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