# Is this series bounded?

1. Mar 24, 2010

### emptyboat

But I can't answer to it.
$$\sum^{\infty}_{1}(-1)^n*(1+\frac{1}{n})^n$$
Is this series bounded?
I can't do anything about that.

Last edited: Mar 24, 2010
2. Mar 24, 2010

### jbunniii

Do you know what

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

equals?

3. Mar 24, 2010

### emptyboat

I know this series diverge by lim an=e, not zero.
But I can't figure out it's bounded.

Last edited: Mar 24, 2010
4. Mar 24, 2010

### hamster143

The sum of any two consecutive terms goes to zero as O(1/n^2). Since the sum of 1/n^2 is convergent, you can prove that your original series is bounded.

5. Mar 24, 2010

### emptyboat

hmm... Do you mean (1 + 1/(n+1))^(n+1) - (1 + 1/n)^n ≤1/n^2?

Last edited: Mar 24, 2010
6. Mar 24, 2010

### hamster143

Correct. You need to prove that there's such N and C that, for all n>N,

(1 + 1/(n+1))^(n+1) - (1 + 1/n)^n ≤ C/n^2

You can do that if you write

(1+1/n)^n = exp(n log (1+1/n))

and then use Taylor expansion of log.

Last edited: Mar 25, 2010
7. Mar 25, 2010

### emptyboat

I computed the process you have given.
log(x)=(x-1) - 1/2*(x-1)^2 + 1/3*(x-1)^3 - ...(correct?)
So My result is
exp (1 - 1/2(n+1) + 1/3(n+1)^2 - ...)
- exp (1 - 1/2n + 1/3n^2 - ...)
let exp(A(n+1)) - exp((An))
exp does not linear, I can't do further.
Help me more...

8. Mar 25, 2010

### hamster143

$$exp(1 - \frac{1}{2n} + \frac{1}{3n^2} + ...) \approx e*( - \frac{1}{2n} + \frac{1}{3n^2}) + ( - \frac{1}{2n} + \frac{1}{3n^2})^2/2 + ...) = e*(-\frac{1}{2n}+\frac{1}{3n^2} + \frac{1}{8n^2} + ...) = e*(-\frac{1}{2n}+\frac{11}{24n^2} + ... )$$

where ... are terms that go down as 1/n^3 or faster.

$$|\frac{1}{2(n+1)} - \frac{1}{2n}| = \frac{1}{2n(n+1)} < \frac{1}{2n^2}$$

9. Mar 25, 2010

Oh I see!!!