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Problem :

Find the laurent expansion of $\frac{1}{ln(1+z)}$ around $z=0$ and find the region of convergnce.

$\frac{1}{ln(1+z)} = \frac{1}{z}\frac{1}{1-(z/2 - z^2/3+z^3/4+...)}$

Maybe the solution comes by using $\frac{1}{1-z}=1+z+z^2+...$ when $|z|<1$

But I don't know whether $|z/2 - z^2/3+z^3/4+...|<1$ or not.

I know that $|z/2 - z^2/3+z^3/4+...|<1$ iff $|\frac{ln(1+z)}{z}-1|<1$

But how can I proceed now?

And I guess the region of convergence is $0<|z|<1$ but I have no confidence.

I really appreciate your help. Thanks.

EDIT : I realized that if $0<|z|<1$, $|z/2 - z^2/3+z^3/4+...|<1$ is not always true, because when $z\rightarrow-1$ with $z$ being a real number, the absolute value is larger than $1$. I'm more clueless now.

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    $\begingroup$ If $f(z) = \sum_{n=0}^\infty a_n z^n$ converges and doesn't vanish for $|z| < r$ then (for $|z| < r$) $\frac{1}{f(z)} = \sum_{n=0}^\infty b_n z^n$ where the coefficients $b_n$ satisfy the recurrence implied by $(\sum_{k=0}^\infty a_k z^k)(\sum_{m=0}^\infty b_m z^m) =1$. Here $f(z) = \frac{\log(1+z)}{z}$ and $r = 1$. $\endgroup$ – reuns Jul 16 '17 at 23:41
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If you are just looking for the Laurent expansion, start with the Taylor series $$\log(1+z)=z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+\frac{z^5}{5}-\frac{z^6}{6}+O\left(z^7\right)$$ $$\frac{1}{\log(1+z)}=\frac{1}{z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+\frac{z^5}{5}-\frac{z^6}{6}+O\left(z^7\right)}$$ and perform the long division to get $$\frac{1}{\log(1+z)}=\frac{1}{z}+\frac{1}{2}-\frac{z}{12}+\frac{z^2}{24}-\frac{19 z^3}{720}+\frac{3 z^4}{160}-\frac{863 z^5}{60480}+\frac{275 z^6}{24192}+O\left(z^7\right)$$

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  • $\begingroup$ With the long division, I got $C_n$ inductively. (I mean, $C_0 = 1$, $C_1 = 1/2$, $C_2 = -1/12$, ... as your solution describes.) But is there any method to write $C_n$ straight, not inductively? (Inductive writing : $C_n = (1/2)C_{n-1}-(1/3)C_{n-2}+...$) $\endgroup$ – Arbitrary Jul 17 '17 at 9:30
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You can find $\lim \limits_{z \to 0} \frac{ln(1+z)-z}{z}$ which comes out to be $0$. Hence $\lvert\frac{ln(1+z)}{z} - 1\rvert$ approaches $0$ when $z$ is close to $0$

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For an explicit expression note that $$ \operatorname{li}(z)=\int\frac{\mathrm dz}{\log z}. $$ Then, according to the Wolfram functions site one has the following expansion about $z=1$. $$ \operatorname{li}(z)=\frac{1}{2}\left(\log(z-1)-\log\left(\frac{1}{z-1}\right)\right)+\gamma+\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)!}(1-z)^{k+1}\sum_{j=1}^{k+1}\frac{B_jS_k^{(j-1)}}{j}, $$ where $B_n$ is the $n$th Bernoulli number and $S_n^{(m)}$ are the signed Stirling numbers of the first kind. Differentiating w.r.t. $z$ then yields $$ \frac{1}{\log z}=\frac{1}{z-1}-\sum_{k=0}^\infty\sum_{j=1}^{k+1}\frac{B_jS_k^{(j-1)}}{k!\, j}(z-1)^k, $$ or equivalently $$ \frac{1}{\log (z+1)}=\frac{1}{z}-\sum_{k=0}^\infty\sum_{j=1}^{k+1}\frac{B_jS_k^{(j-1)}}{k!\, j}z^k. $$ You can verify that this gives the same answer provided by Claude Leibovici.

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