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Ean Hughes

Answered question

2022-04-16

How do we go from

\frac{1}{1 + (\frac{x}{2}) + (\frac{x^{2}}{3}) + (\frac{x^{3}}{4}) + \dots}

Answer & Explanation

cab65699m

Beginner2022-04-17Added 14 answers

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.
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 the 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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