Find $n$ such that $365 \left(1-\left(\frac{364}{365}\right)^n - n\frac{364^{n-1}}{365^n} \right) > 1$

Question

I need to find $n$ such that $365 \left(1-\left(\frac{364}{365}\right)^n - n\frac{364^{n-1}}{365^n} \right) > 1$. The answer is $n \ge 28$. Expanding the equation out, rearranging it, taking logarithm, I get $$\log(364) > \log(365) + n\log(364/365) + \log(n) +(n-1)\log(364/365).$$

I don't know what to do next. I would appreciate if you give me some hint.

Answer 1

I don't get the same inequality as you do. We have $$ 365 \left(1-\left(\frac{364}{365}\right)^n - n\frac{364^{n-1}}{365^n} \right) > 1\\ 364> \frac{364^n}{365^{n-1}}+n\left(\frac{364}{365}\right)^{n-1}\\ 364>(364+n)\left(\frac{364}{365}\right)^{n-1}\\ \log364>\log(364+n)+(n-1)\log\left(\frac{364}{365}\right)\\ (n-1)\log\left(\frac{365}{364}\right)>\log\left(\frac{364+n}{364}\right)\\ (n-1)\log\left(1+\frac1{364}\right)>\log\left(1+\frac n{364}\right) $$ I haven't carried it past this point. I would try applying the inequalities $$x-\frac{x^2}{2}<\log(1+x)<x,\ |x|<1$$

Answer 2

Consider the inequality writing $a=364$, $$(a+1)\left(1-\left(\frac{a}{a+1}\right)^n-n\left(\frac{a^{n-1}}{(a+1)^n}\right)\right)>1.$$ A bit of rearranging leads to $$\left(\frac{a+1}{a}\right)^{n-1}>1+\frac{n}{a}.$$ Take the natural log of both sides to get $$ (n-1)\ln \left(\frac{a+1}{a}\right) > \ln\left(1+\frac{n}{a}\right).$$ The simplest way to go from here is to use the inequality $\ln(1+x)\ge \frac{2x}{2+x}$ for all $x \ge 0$. As $\frac{n}{a}$ is small it is reasonably accurate. Substituting in that inequality and rearranging eventually yields $$ n^2+b\,n-2a>0 $$ where $$b=2a-1-\frac{2}{\ln\left(\frac{a+1}{a}\right)}.$$ Thus $$ n>-b+\sqrt{b^2+2a}\approx 27.9998. $$

However, this is a necessary condition but possibly not sufficient. To find the exact value of $n$ I think you would require a numerical method. You can getter a slightly better bound by instead starting from $$\left(\frac{a+1}{a}\right)^{n-2}>1+\frac{n-1}{a+1}.$$ Using the same log inequality and working it through it is found that $n$ has to be slightly larger than $28$.

This is confirmed when checking the original problem. It does not hold for $n=28$, but does hold for $n=29$. So if $n$ is an integer one requires $n\ge 29$ not $n\ge 28$.

Perhaps you could look into the log inequality used to see if you could put a bound on the error induced by using it, or find a better inequality, so that you have a better idea of the bound on $n$.