How does one solve the following inequality for $n$, without trial and error, and assuming $n$ can only be an integer?
The inequality is $(n+1)!-1>10^9$.
I want to find the minimum value of $n$ such that $(n+1)!-1>10^9$. How does one do this without graphing the inequality, or using a calculator? I figured it out throwing trial and error on a calculator, but I desire a more elegant solution, one that gets to the answer algebraically, and without a calculator or graph. Any suggestions?
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If you know that $10!$ is between $3$ and $4$ million, then it seems that you would need to go three terms further in the factorial sequence (multiplying by $11$, $12$, $13$) to first exceed $1$ billion.
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Observe that between $1$ and $10$, multiplying $1$ and $10$, $2$ and $9$, etc., every pairwise product contributes at least a factor of $10$. Thus $10!$ is at least $10^5$ (but of course you can deduce quite easily that it must actually be at least $10^6$). Then multiplying by $11$ and $12$ contributes about $10^2$. Now we know that $12!>10^8$. So the largest possible value of $n+1$ is now $13$, and you only have to try $12!$ to show that it is indeed smaller than $10^9$. Minimal brute force involved!
A possible way would be to use upper and lower bound to the factorial function. There are of course very precise ones, but you said you don't want to use a calculator, so we take simple ones:
$$\left(\frac{n}{2}\right)^{\frac{n}{2}}\leq n!\leq n^n$$
(I leave the simple proofs to you, as they are quite insightful).
Then we first want to solve the easier inequation $m!>10^9$. We immediately see that we must have $m>9$. This already gives you some starting point. An upper bound on the other hand is 20. It is still somewhat bruteforce.
A better approximation is $n!\approx\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ which will help you more to get a good starting guess.