Limit of sequence

1. Nov 30, 2004

twoflower

Hi all,

I can't move with this limit:

$$\lim_{n \rightarrow \infty} n \left( \left( 1 + \frac{1}{n} \right)^{n} - e \right)$$

Could someone help me please? Or some hint...

Thank you

2. Nov 30, 2004

e(ho0n3

Find out what

$$\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^{n}$$

is. It should all be clear from then on.

3. Nov 30, 2004

twoflower

Well I know it is equal to e, but then

$$\lim_{n \rightarrow \infty} n \left( e - e \right) = \lim_{n \rightarrow \infty} n \left ( 0 \right) = 0.\infty$$

Which is undefined...

4. Nov 30, 2004

shmoe

Are you opposed to using the series expansions of log(1+x) and e^x? You can re-write the (1+1/n)^n term as e^(n*log(1+1/n)). Use the Taylor series for log(1+x), at least 2 terms +remainder. Then you can factor out an e and use the series for e^x, (2 terms+remainder here as well), and that will do it.

If you are opposed to using series but have learned them, it's probably possible to modify the above into a squeeze limit type of proof. If you haven't learned Taylor series yet, I'll have to think of a more elementary method, though it will probably be a thinly disguised version of the above.

5. Nov 30, 2004

twoflower

Thank you shmoe, maybe this limit is really supposed to be solved using Taylor series, because it is from sample calculus test we're gonna take, but as I see in sylabus, we'll learn Taylor series before the test. It confused me, because the sequences are already behind us and I thought I should already be able to solve any limit of sequence....

6. Nov 30, 2004

stunner5000pt

you cannot sub in the value for e in between the limit like that

7. Nov 30, 2004

stunner5000pt

i wonder if this will work though

split the limit so you have the limits of n(1-1/n)^n and n e subtracted from each other.

now the limit of n (1-1/n)^n is infinity while the limit of n E is also infinity

however the first one converges to e more slowly that E already does

so n E >n(1-1/n)^n and the limit is negative infinity

8. Nov 30, 2004

twoflower

The limit should be -e/2...

9. Nov 30, 2004

stunner5000pt

check what you entered into maple/mathematica/matlab... it isn't -e/2, it is -infinity

10. Nov 30, 2004

shmoe

No, this won't work. You're essentially saying infinity-infinity=-infinity because the first infinity is getting there slower. This is bunk. An indeterminate infinity-infinity limit form can potentially equal anything we like(-e/2 is correct in this case, done by hand with the method I suggested).

twoflower-series is likely what's intended then. I can't see another way that isn't unnecessarily complicated.

11. Nov 30, 2004

NateTG

If you can show that
$$-\frac{e}{2n} - \epsilon_1(n) \leq \left( 1 + \frac{1}{n} \right)^{n} - e \leq -\frac{e}{2n} + \epsilon_2(n)$$

Where $\epsilon(n)$ is some expression that goes to zero faster than
$\frac{1}{n}$ then you're set, so perhaps you should look at the convergence of $\left( 1 + \frac{1}{n} \right)^{n}$

12. Nov 30, 2004

twoflower

I tried it now. Maple gives -e/2

13. Nov 30, 2004

shmoe

Nate, do you have a simple method in mind for those inequalities?

The obvious approach to me involves some bounds on log(1+x) and e^x that I would prove using series.

14. Nov 30, 2004

twoflower

Ok, so you say guys that the most natural way to find this limit is to use Taylor series. If it's right, I will skip this one. Or does anyone have simplier approach to show?