# Homework Help: Limit of sequence

1. May 12, 2010

### Malmstrom

1. The problem statement, all variables and given/known data
Prove that
$$\lim_{n \rightarrow \infty} \sqrt[n]{\frac{2^n}{n!!}}=0$$

2. Relevant equations

3. The attempt at a solution
Seems really tricky ...

2. May 12, 2010

### Fragment

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

-F

3. May 12, 2010

### Malmstrom

Not sure what you mean with exponential form.

4. May 12, 2010

### Malmstrom

If you mean

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

it's not helping me.

5. May 12, 2010

### 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

6. May 12, 2010

### Malmstrom

It's a little simplified, as I get

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

7. May 12, 2010

### Char. Limit

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

8. May 12, 2010

### 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...

Last edited: May 12, 2010