Finding $\lim_{n\to \infty}(n!-10^n)$

Question

$$\lim_{n\to \infty}(n!-10^n)$$

So of course $10^n<<n!$ but how can we prove that the limit is $\infty$?

score 3 · Answer 1 · edited Feb 12 '20 at 13:14

3

You can write the sequence as $n!\left(1-\frac{10^n}{n!}\right)$, then the result simply follows from arithmetic of limits.

edited Feb 12 '20 at 13:14

David C. Ullrich

92,839

answered Feb 12 '20 at 12:43

Mark

43,582

We can not evaluate $lim( n!)\cdot lim (1-\frac{10^n}{n!})$ – newhere Feb 12 '20 at 12:50
The limit $\lim_{n \to \infty}(n!)$ does not converge. – tajiri_numero_1 Feb 12 '20 at 12:59
2

$n!$ tends to $\infty$ while $1-\frac{10^n}{n!}$ tends to $1$. Hence the product tends to $\infty$. – Mark Feb 12 '20 at 13:11

score 3 · Accepted Answer · answered Feb 12 '20 at 13:00

3

Hint:

For $m>10$,

$$(m+n)!>m!\,10^n.$$

Find $m$ such that $m!>10^m$.

answered Feb 12 '20 at 13:00

Sarvesh Ravichandran Iyer · Answer 3 · 2022-10-07T13:00:37.773

Write $n! -10^n = 10^n \left(\frac {n!}{10^n} - 1 \right)$.

Now, $10^n \to \infty$. Let $b_n = \frac {n!}{10^n} - 1$. Then, $b_n+1 = \frac{n!}{10^n}$, and $b_{10}+1>0$ obviously.

Observe that $\frac{b_{n+1}+1}{b_n+1} = \frac{n+1}{10} \geq \frac{11}{10}$ for $n\geq 10$.

Therefore, $b_{n} \geq \left(\frac{11}{10}\right)^{n-10} \left(b_{10}+1\right)-1$ for $n \geq 11$. Thus, by domination, we see that $b_{n} \to +\infty$ as $n \to +\infty$.

Finally, $n!-10^n = 10^nb_n \to +\infty$.

score 1 · Answer 4 · answered Feb 12 '20 at 13:45

Starting from @Yves Daoust's answer, you need to solve for $n$ the equation $$n!=10^n$$ If you look at this question of mine, asking for the approximate solution of $$n!=a^n\, 10^k$$ you will find a magnificent approximation provided by @robjohn, an eminet user on this site.

Making $k=0$ and $a=10$, the answer is $$n \sim 10\,e \exp{\left(W\left(-\frac{\log (20 \pi )}{20 e}\right) \right) }-\frac 12$$ where $W(.)$ is Lambert function.

Since the argument is quite small $\left(\frac{\log (20 \pi )}{20 e} \approx 0.076\right)$, you can use the series expansion $$e^{W(x)}=1+x-\frac{1}{2}x^2+\frac{2 }{3}x^3+O\left(x^4\right)$$ to get $n=24.5257$ then $n=25$.

Numerically, the exact solution of the equation is $24.5264$ (!!)

score 0 · Answer 5 · answered Feb 12 '20 at 13:34

Yet another approach, inspired by the one of Yves Daoust. Assume $n > 100$ and write $n=100+m$. Then $n! > 100^m = 10^{2m}$. So in that case $$n! - 10^n > 10^{2m} - 10^{100+m} = 10^m \cdot (10^m - 10^{100}).$$ Now as we let $m$ tend to infinity, we observe that both terms of the product tend to infinity.

Peter Szilas · Answer 6 · 2020-02-12T15:02:12.557

0

$$e^{10}= 1+10/1!+10^2/2!+\cdots + 10^k/k!+ \cdots$$

Since this series is convergent, it follows that $\lim_{k \rightarrow \infty}10^k/k!=0$.

Hence, for sufficiently large $k:$ $10^k/k! <1/2$.

$k!(1-10^k/k!)>k!(1-1/2)=(1/2)k!$.

Another attempt:

Note: $k!>(k/2)^{k/2}$; for $k >2$.

$f(k):= k!(1-10^k/k!)$;

$f(k) >k!(1-\dfrac{100^{k/2}}{(k/2)^{k/2}})=$

$k!(1-(\dfrac{100}{k/2})^{k/2})$.

Take the limit.

edited Feb 12 '20 at 15:02

answered Feb 12 '20 at 13:41

Peter Szilas

21,123

Gary.Thanks, looks nice. – Peter Szilas Feb 12 '20 at 13:47

Finding $\lim_{n\to \infty}(n!-10^n)$

6 Answers6