Proving Limit of Sequence: $\sqrt[n]{\frac{2^n}{n!}}=0$

[Malmstrom]

Homework Statement

Prove that
$$\lim_{n \rightarrow \infty} \sqrt[n]{\frac{2^n}{n!}}=0$$

Homework Equations

The Attempt at a Solution

Seems really tricky ...

 

[Fragment]

You could try rewriting it in exponential form, getting rid of the root. This will help you see what to do next.



-F

 

[Malmstrom]

You could try rewriting it in exponential form, getting rid of the root. This will help you see what to do next.



-F

Not sure what you mean with exponential form.

 

[Malmstrom]

If you mean

$$e^{\frac{1}{n} \log (\frac{2^n}{n!}) }$$

it's not helping me.

 

[Fragment]

How about, $$\sqrt[n]{x}$$=$$x^{n^{-1}}$$

This way you might not need logarithms, and you might be able to simplify it enough for it to be clear to you.


-F

 

[Malmstrom]

It's a little simplified, as I get

$$2 \sqrt[n]{\frac{1}{n!}}$$

.. but what about this term?

 

[Char. Limit]

Well, what happens to that term as $n -> \infty$?

 

[Malmstrom]

It hopefully goes to zero but this is not a proof. It is not clear at all because for instance

$$\lim_{n \rightarrow \infty} \sqrt[n]{\frac{1}{n}} = 1$$

so...