Taylor series of $\frac{1} {\ln (t+1)}$ at $t=0$.

Question

I have tried to use the natural Taylor expansion of $\ln(t+1)$ and working with long division but I get the result $\frac{1}{t}+\frac{1}{2}+\frac{t}{12}-\frac{t^2}{24}+\cdots$ instead of $\frac{1}{t}+\frac{1}{2}-\frac{t}{12}+\frac{t^2}{24}+\cdots$ . It seems that after $\frac{1}{2}$ the signs are changing in wrong order. I have tried to do this several time but still don't works. Can you show me your calculations?

Answer 1

Using the Binomial series for $(1-x)^{-1}$, $\displaystyle \frac{1}{\ln(t+1)}=\big(t(1-\tfrac{1}{2}t+\tfrac{1}{3}t^2-\ldots)\big)^{-1}=\frac{1}{t}\big((1-(\tfrac{1}{2}t-\tfrac{1}{3}t^2+\ldots)\big)^{-1}$ $\displaystyle =\frac{1}{t}\big((1+(\tfrac{1}{2}t-\tfrac{1}{3}t^2+\ldots)+(\tfrac{1}{2}t-\tfrac{1}{3}t^2+\ldots)^2 +\ldots\big)$ $\displaystyle =\frac{1}{t}\big(1+\tfrac{1}{2}t-\tfrac{1}{3}t^2+\tfrac{1}{4}t^2 +\ldots\big)$ $=\frac{1}{t}+\tfrac{1}{2}-\tfrac{1}{12}t+\ldots$