# Maclaurin series for $\frac{x}{e^x-1}$

Maclaurin series for

$$\frac{x}{e^x-1}$$

$$1-\frac x2 + \frac {x^2}{12} - \frac {x^4}{720} + \cdots$$

How can i get that answer?

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And, I don't want to take derivative, because it is going to take too much time. –  Bek Abdik Feb 23 '13 at 8:09
Do you mean the $5$ terms only? –  André Nicolas Feb 23 '13 at 8:21
Yes, I need to find untill x^4's term –  Bek Abdik Feb 23 '13 at 8:26
For the whole series, see en.wikipedia.org/wiki/Bernoulli_number#Generating_function –  Hans Lundmark Feb 23 '13 at 8:33
I was typing it out, but Ittay Weiss is faster. It will work very nicely for the first few coefficients. –  André Nicolas Feb 23 '13 at 8:35

Let $f(x)=\frac {x}{e^x-1}$ and consider the product $(e^x-1)\cdot f(x)=x$. Since $f$ is infinitely differentiable it follows that it has a Taylor series. Now, the Taylor series for $e^x-1$ is $$\sum_{k=1}^\infty \frac{x^k}{k!}$$(obtained immediately from the Taylor series for $e^x$). Thus, if $$\sum _{k=0}^\infty c_kx^k$$ is the Taylor series for $f(x)$ then $$(\sum _{k=1}^\infty \frac {x^k}{k!})\cdot (\sum _{k=0}^\infty c_kx^k)=x$$ and dividing by $x$ yields $$(\sum _{k=0}^\infty \frac {x^{k}}{(k+1)!})\cdot (\sum _{k=0}^\infty c_kx^k)=1,$$ from which, by expanding and equating coefficients, we obtain

$1/1!\cdot c_0 = 1$,

$1/1!\cdot c_1 + 1/2!\cdot c_0 = 0$

$\vdots$

$\sum _{j=0}^m\frac{1}{(j+1)!}\cdot c_{m-j}$

$\vdots$

You solve these equations inductively to obtain the values for the $c_k$.

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One way is to write $e^x-1$ as $1 + x + x^2/2 + ... - 1$ and then factor out $x$ and cancel up the top and expand it as geometric series and collect the coefficients of like powers.

\displaystyle \begin{align*} e^x - 1 &= x + \frac{x^2}{2!} + \frac{x^3}{3!} + o(x^4)\\ \frac{x}{e^x - 1} &= \frac{1}{1 + \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right )} \\ &= 1 - \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right ) + \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right )^2 - \left( \frac x 2 - \frac{x^2}{6} + o(x^3) \right )^3 + \left( \frac x 2 + \frac{x^2}{6} + o(x^3) \right )^4... \\ &= 1 - \frac{x}{2} + x^2 \left( \frac 1 4 - \frac 1 6 \right ) + x^3 \left(-\frac{1}{4!} + 2 \cdot \frac 1 2 \cdot \frac 1 6 - \frac{1}{2^3} \right ) + x^4 \left(-\frac{1}{5!} + \frac{1}{6^2} + 2 \cdot \frac 12 \cdot \frac{1}{4!} + \frac{1}{2^4}\right )+o(x^5) \end{align*}

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I think you missed a $-$ in the $^3$ term. –  Alyosha Feb 23 '13 at 13:06
@Alyosha thanks edited :) –  Santosh Linkha Feb 23 '13 at 14:59