More on Sequences — Question 9

*Question 9 For $n\ge2$, let $a_n=\left(1-\dfrac1n\right)^{n^2}$.

1. Take logarithms and use $\ln(1-x)\le-x$ for $0<x<1$ to obtain a useful bound.

2. Prove that $a_n\to0$.

3. Use $\ln(1-x)=-x-x^2/2+O(x^3)$ to describe the rate of decay more precisely.

*Question 9 – Solution

For $n\ge2$, $0<1-1/n<1$, so $a_n>0$ and logarithms are valid: $$\ln a_n=n^2\ln\left(1-\frac1n\right).$$

For $0<x<1$, $\ln(1-x)\le-x$. With $x=1/n$, this gives $\ln a_n\le-n$. Because the exponential function is increasing, $$0<a_n\le e^{-n}\longrightarrow0.$$ The Squeeze Theorem proves $a_n\to0$.

For a sharper description, substitute $x=1/n$ into $\ln(1-x)=-x-x^2/2+O(x^3)$: $$\ln a_n=n^2\left(-\frac1n-\frac1{2n^2}+O(n^{-3})\right) =-n-\frac12+O(n^{-1}).$$ Exponentiating both sides gives $$a_n=e^{-n-1/2+O(1/n)}\sim e^{-n-1/2}.$$ The notation $u_n\sim v_n$ means $u_n/v_n\to1$. Thus the quadratic exponent repeats a base close to $1$ often enough to produce decay essentially like $e^{-n}$, with the additional constant factor $e^{-1/2}$.