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Use the basic null sequences to show that $\left(\dfrac{n^{10}10^n}{n!}\right)_{\large n\in \mathbb{N}}$ is a null sequence.

Will someone provide a hint?

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"Let $\varepsilon>0$ be given..." – andybenji Jan 10 '13 at 19:09
@andybenji But the question indicates not to work from first principles, but instead to use basic null sequences to prove the statement. – Ryan Jan 10 '13 at 19:12
@Ryan I think he meant it as a joke. – Git Gud Jan 10 '13 at 19:15
@Ryan Yeah, factorials are very hard to work with using the definition. Can you think of a sequence that's greater in absolute value (after finitely many terms) than the given one but is still a null sequence? – andybenji Jan 10 '13 at 19:25
@Ryan Not at all. Let $a_n>0$. If $\lim a_{n+1}/a_{n}<1$ then $\lim a_n=0$. What's more, $\sum a_n$ converges. – Pedro Tamaroff Jan 10 '13 at 22:29
up vote 2 down vote accepted

For large enough $n$, $\Bigl(\frac{n^{10}10^n}{n!}\Bigr)$ is less than $\Bigl(\frac{11^n}{n!}\Bigr)$ because $1.1^n$ is eventually larger than $n^{10}$ (to prove this take logarithms). So we only need to show that $x_n = \Bigl(\frac{11^n}{n!}\Bigr)$ is a null sequence and then an application of the Sandwich Theorem yields the result.

To this end, firstly note that $x_n$ > 0 for all n and so for n = 22 your sequence is equal to some constant greater than 0. Call this constant c.

Show that for n > 22, $\Bigl(\frac{x_{n+1}}{x_{n}}\Bigr) < 1/2$ and so, roughly speaking, getting from $x_n$ to $x_{n+1}$ requires you multiplying by a number less than $1/2$.

Then write $x_n < c \Bigl(\frac{1}{2^n}\Bigr)$ for all n > 22, and as mentioned above apply the Sandwich Therorem.

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Thanks Adam. The key observation is that $1.1^n$ is eventually larger than $n^{10}$. Taking logarithms, this turns out to be clearly untrue, doesn't it? – Ryan Jan 10 '13 at 20:08
Well let me ask you: What is log $1.1^n$ ? What is log$(n^{10})$ ? – Adam Rubinson Jan 10 '13 at 20:12
right, and log1.1 > 0, so that (10logn)/(log1.1) > 0, and also logn is much much slower than n as n tends to infinity. Working in base 10 for example, log(1000000) = 6 < 1000000. You can formally prove that lim(log(x) / x) = 0 if you wanted to, but I assumed that this was known. – Adam Rubinson Jan 10 '13 at 20:36
Check your original assertion "$1.1^n$ is eventually greater than $n^{10}$". This is equivalent to $\frac{log (1.1)}{10} > \frac{log(n)}n$, which is false, right? – Ryan Jan 10 '13 at 20:45
@Ryan Not false at all. You have a (positive) constant on the left hand side, and something that tends to $0$ as $n\to\infty$ on the right hand side. – mrf Jan 10 '13 at 22:21

For $n$ sufficiently large $n^{10}<10^n$. Now consider:

$$\frac{100^n}{n!}=\frac{100}{n}\cdot \frac{100}{n-1}\cdots \frac{100}{1}$$

For all $n>100:\;\;\dfrac{100}{n}\leq\dfrac{100}{101}$ so from the above: $\dfrac{100^n}{n!}\leq N\cdot\left(\dfrac{100}{101}\right)^{n-100}$ which yields

$$\dfrac{10^n n^{10}}{n!}<N\cdot\left(\dfrac{100}{101}\right)^{n-100}\;\;\text{etc.}$$

Otherwise, a nice way is to consider:

$$e^{100}=\sum_0^{\infty} \frac{100^n}{n!}$$

$\to$ series converges so the summand must be a null-sequence.

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