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How do we go from

\begin{equation*} \frac{1}{1+(\frac{x}{2})+(\frac{x^2}{3})+(\frac{x^3}{4})+\dots} \end{equation*}


\begin{equation*} 1-(\frac{x}{2})-(\frac{x^2}{12})-\dots? \end{equation*}

If I could consolidate the series in denominator in some form, then I could use binomial expansion in numerator by raising it to power $-1$, but I don't see the one in denominator in any known form other than converting it to $\log.$

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In the denominator of what? In the numerator of what? – Mariano Suárez-Alvarez Sep 25 '10 at 21:50
I mean the first series is in denominator of 1. – Pupil Sep 25 '10 at 21:54
If it's helpful for anyone else, I understand the question to be asking for an explanation of $$\frac{1}{\sum_{k=0}^{\infty}\frac{x^k}{k+1}}=1-\frac{x}{2}-\frac{x^2}{12}-\fra‌​c{x^3}{24}-\frac{19x^4}{720}-\frac{3x^5}{160}-\frac{863x^6}{60480}-\cdots$$ – Isaac Sep 25 '10 at 22:15
@Isaac: I interpreted it the same way too and added an answer with that interpretation. – Aryabhata Sep 25 '10 at 22:17
Well, I was not very subliminally asking for clarification in the question itself... – Mariano Suárez-Alvarez Sep 25 '10 at 22:44
up vote 6 down vote accepted

Using $$-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$$

We have that

$$ - \frac{x}{\log(1-x)} = \frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots}$$

Now the series for $$\frac{x}{\log(1-x)}$$ is well known.

See the question asked on this very site here:

And the page here: (search the page for Gregory).

The series expansion is given by

$$\frac{x}{\log(1-x)} = \sum_{k=0}^{\infty} C_{k} x^{k} = -1 + \frac{x}{2} + \frac{x^2}{12} + \frac{x^3}{24} + \dots$$

The $C_{k}$ are called as Gregory coefficients. The wiki page I linked above tells you how they can be calculated using a recursive formula.

So to answer your question, we get

$$\frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots} = - \sum_{k=0}^{\infty} C_{k} x^k = 1 - \frac{x}{2} - \frac{x^2}{12} - \frac{x^3}{24} - \dots$$

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@Harpreet: Use the dollar signs and latex between them. Example $x^2+y^2=z^2$ looks like $x^2+y^2=z^2$ – Aryabhata Sep 25 '10 at 22:23

Let $f(x) = \sum f_n x^n$ be a formal power series. To compute the inverse of $1 - x f(x)$, write

$$\frac{1}{1 - x f(x)} = \sum_{n \ge 0} x^n f^n(x).$$

To compute the coefficient of $x^k$ it suffices to compute the contributions from the first $k$ terms on the RHS, so this results in a finite algorithm which will let you compute any particular coefficient of the inverse. If $f$ has special properties (for example it is a polynomial or more generally meromorphic) then often one can even find a nice closed form.

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