Given $f(x)=\sum_{n=0}^{\infty}\frac{1}{\sqrt{1+n}}x^n$ find the first three coefficients of the power series $\frac{1}{f(x)}=\sum_{n=0}^{\infty} b_n x^n$.
I tried the following: rewrote to $(1+n)^\frac{-1}{2}$ then used binomial series. So I got $1-\frac{1}{2}x-\frac{1}{16}x^2$. Then the fraction so $1-2x-16x^2$. But I sense this cannot be correct. Could anyone provide a helpful insight?
On
$$1={f(x)\over f(x)}=\left(1+{1\over\sqrt2}x+{1\over\sqrt3}x^2+\cdots\right)\left(b_0+b_1x+b_2x^2+\cdots\right)\\ =b_0+\left(b_1+{b_0\over\sqrt2} \right)x+\left(b_2+{b_1\over\sqrt2}+{b_0\over\sqrt3} \right)x^2+\cdots$$
implies $b_0=1$, $b_1+{b_0\over\sqrt2}=0$, and $b_2+{b_1\over\sqrt2}+{b_0\over\sqrt3}=0$, hence $b_1=-{1\over\sqrt2}$ and $b_2={1\over2}-{1\over\sqrt3}$, so
$${1\over f(x)}=1-{1\over\sqrt2}x+\left({1\over2}-{1\over\sqrt3} \right)x^2+\cdots$$
The simplest method consists in the division by increasing powers of $1$, up to the chosen , of the expansion of $f(x), truncated at the same order.
You should obtain, if I'm not mistaken, $$1-\frac x{\sqrt2}+\Bigl(\frac12-\frac1{\sqrt 3}\Bigr)x^2+o(x^2).$$