Implication from an asymptotic equation.

Question

Let $(z_n)$ be some zero sequence, $\lim_{n\to\infty}z_n=0$.

Assume the equation $$ (n-1)\cdot z_n+O(n\cdot z_n^2)+\log(z_n)=0\text{ as }n\to\infty. $$ holds.

Show that this implies $\lim_{n\to\infty}n\cdot z_n^2=0$.

Can you help me?

Answer 1

Let $f(n)=O(nz_n^2)$, that is $|f(n)|\le C|nz_n^2|$ for sufficiently large $n$.

Let $r_n = (n-1)z_n + f(n) + \log(z_n)$. Multiplying by $z_n$ and solving for $nz_n^2$ yields:

$$ n z_n^2 = r_nz_n + z_n^2 - z_n\log(z_n) - z_nf(n) $$

Taking the absolute values and applying the triangle inequality then gives

$$|nz_n^2| \le |r_nz_n+z_n^2 - z_n\log(z_n)| + |z_n| \cdot C |nz_n^2| $$

Hence

$$|nz_n^2| \le \underbrace{\frac{1}{1-|z_n|C}}_{\to 1}|\underbrace{r_nz_n}_{\to 0}+\underbrace{z_n^2}_{\to 0} - \underbrace{z_n\log(z_n)}_{\to 0}| \longrightarrow 0$$