This question was asked in my quiz (now over) and I was unable to solve it.
For $0<y<1$ , prove that $\log[(1-y) ^{-1}] = y + O(y^2)$.
I tried using expansion of $\frac{ 1} { 1-y} $ and trying to prove $\frac{ log ( 1+ y+ y^2+...) -y }{y^2}$ will always be less than a particular constant for all $y \geq a$, but unable to do so.
So, can you please help with this.
Apologies for the typo.
On
Hint:
Considering natural logarithm
$$\ln\left(\frac{1}{1-y}\right) = \ln\left(1+\frac{y}{1-y}\right)=\\ =\frac{y}{1-y}-\left(\frac{y}{1-y}\right)^2\frac{1}{2}+ \left(\frac{y}{1-y}\right)^3\frac{1}{3}-\cdots=\\ \text{Addition}\\ =y+\frac{y^2}{1-y} - \left(\frac{y}{1-y}\right)^2\frac{1}{2}+ o(y^2)=y+O(y^2)$$
To elaborate on other answers and comments, note first that the claim is false as $y$ approaches 1, since $$ \lim_{y\to 1} \left[ \log((1-y)^{-1}) \right] = \infty. $$ That is, the claim is false if $y$ is allowed to be arbitrarily close to 1. Thus we need a uniform bound on the possible values of $y$; i.e. we need to assume the slightly stronger statement that $y\leq 1-\delta$ for some fixed positive number $\delta$.
Now the result follows almost immediately from the Taylor series expansion of $-\log(1-y)$: $$ -\log(1-y) = y + \frac{y^2}{2} + \frac{y^3}{3} + \cdots. $$ The tail is bounded via $$ \frac{y^2}{2} + \frac{y^3}{3} + \cdots \leq y^2 + y^3 + y^4 + \cdots = \frac{y^2}{1-y}, $$ where we have summed the geometric series since $\left|y\right| < 1$. Then $$ \frac{y^2}{1-y} \leq \frac{y^2}{1-(1-\delta)} = \frac{y^2}{\delta}. $$ Therefore $$ -\log(1-y) = y + O(y^2), $$ where the constant implied in the $O$ notation depends on $\delta$ and may be taken to be $\frac{1}{\delta}$.