Im reading a book in Numerial analysis and I have the following which I dont understand involving inequalities and factorials, What i have is the following:
$$\frac{1}{(2n+1)!(2n+1)} \leq 5*10^{-9}$$
The book gives the answer $2n+1 \leq 11$, so $n \geq5$, but i dont know how the procedure. As an example what would be the value of $n$ so that the following holds: $$\frac{1}{(2n+2)!(2n+1)} \leq 5*10^{-9}$$ Using matlab i found that n $\geq 4$.
$$ f(n) = \frac{1}{(2n+1)!(2n+1)} $$ we know that $$ f(n+1) < f(n) $$ i.e. strictly decreasing function with $n$. thus trying a few values of $n$ we find \begin{align} \text{n=4: } \frac{1}{9!9} &=& \frac{1}{3265920}\approx 3\times10^{-7}\\ \text{n=5: } \frac{1}{11!11} &=& \frac{1}{439084800}\approx 2\times10^{-9}\\ \text{n=6: } \frac{1}{13!1} &=& \frac{1}{80951270400}\approx 1\times10^{-11}\\ \end{align} thus we can conclude that your limit lies between $5 < n <6$ thus the highest integer is $n=5$ you can do the same for the other function.
I preface this with the fact that you may have to use a "more" mathematical approach than just trying values. i.e. I determined by using $n$ but you give a relation $2n+1 \leq 11$ to derive your $n$, so you may have to use a different approach.