# Stirling's approximation problem

1. Mar 27, 2009

### tnutty

1. The problem statement, all variables and given/known data

$$\sum n^n/ n!$$

This example is in the book, it concludes that the above series is :

(1 +1/n)^n, which converges to e and n->infinity

how is this so?

If i take the root test then, is not answer 1 as n->infinity

Can you explain?

2. Mar 27, 2009

### Dick

Re: Divergent/convergent

How about using Stirling's approximation for the n!?

3. Mar 27, 2009

### tnutty

Re: Divergent/convergent

Haven't got there yet

4. Mar 27, 2009

### Dick

Re: Divergent/convergent

That series clearly diverges. Every term is greater than 1. What's the real question? Does it have an x^n in it? Then use the ratio test.

Last edited: Mar 27, 2009
5. Mar 27, 2009

### tnutty

Re: Divergent/convergent

Prove the following:

lim (1 +1/n)^n = e
n->infinity

6. Mar 27, 2009

### Dick

Re: Divergent/convergent

That's not a series. That's a limit. What's your definition of e? Because that's one of them.

7. Mar 27, 2009

### tnutty

Re: Divergent/convergent

so how is it that the above limit goes to e = 2.7128282....

8. Mar 27, 2009

### the_awesome

Re: Divergent/convergent

dont u just re-arrange to find n if e = 2.7?

if u draw the graph on ur calculator u can see that the left side jumps up then stops at (-1.27, 7.15), then all of a sudden on the right side it shows up at about (0.3, 1.57) again, then moves up slowly

Last edited: Mar 27, 2009
9. Mar 27, 2009

### tnutty

Re: Divergent/convergent

10. Mar 28, 2009

### Staff: Mentor

Re: Divergent/convergent

Let y = (1 + 1/n)n (no limits just yet)
ln y = ln (1 + 1/n)n = n ln (1 + 1/n)

lim ln y = lim n ln (1 + 1/n) ( all limits as n --> $\infty$)

= lim ln (1 + 1/n) / (1/n) (the limit on right is of the form 0/0, so L'Hopital's Rule applies)
$$= lim \frac{\frac{-1/n^2}{1 + 1/n}}{\frac{-1}{n^2}}$$
= lim 1/(1 + 1/n) = 1

So lim ln y = 1, or
ln (lim y) = 1, or
lim y = e1 = e

Therefore, lim (1 +1/n)n = e