Does the series converge or diverge ( -1)^n (1-1/n)^n

[Simkate]

I just want to know if what i did is correct...please help Thank YOU!

∑(n =200 to ∞) ( -1)^n (1-1/n)^n = (-1 + 1/n)

This is a alternation Series
therefore the 2 condtions need to be satisfied for it to be Convergent.
i) is true

ii) lim b_n = 0 ?

L= lim (n--> ∞) (1-1/n)^n

= lim ( n--> 0^+) ln (1-t) / t = -1

= lim(n-->0^+) ln (-1/1/t) / (1) = -1

Thus, L=e^-1 = 1/e which not equal to O.

Therefore the series is not absolutely convergent and is divergent


AM I CORRECT?

 

[vela]

ii) lim b_n = 0 ?

L= lim (n--> ∞) (1-1/n)^n

= lim ( n--> 0^+) ln (1-t) / t = -1

= lim(n-->0^+) ln (-1/1/t) / (1) = -1

Thus, L=e^-1 = 1/e which not equal to O.

Therefore the series is not absolutely convergent and is divergent


AM I CORRECT?

Yes and no. Your basic idea is correct, but what you wrote down needs a lot of work. For instance, the original limit is as n goes to infinity, but your expression suddenly has t in it and n is suddenly going to 0⁺. You need to say what substitution you're doing and use only one variable at a time. Next, you write the limit equals -1. How did you get that? And if you already have the answer, why did you do more steps to evaluate the limit after that? You also differentiated incorrectly and then claimed the limit of t as t->0+ is equal to -1. And where did the log come from? Are you really saying the limit of a function and the limit of the log of that function are equal?

 

[Je m'appelle]

$$\lim_{n \rightarrow \infty} (1-\frac{1}{n})^n = L$$

as $$n \rightarrow \infty$$, the expression tends to

$$(1-0)^{\infty} = 1 = L$$

What conclusions do you get when

$$L = 1,\ L \neq 0$$

 

[vela]

$$\lim_{n \rightarrow \infty} (1-\frac{1}{n})^n = L$$

as $$n \rightarrow \infty$$, the expression tends to

$$(1-0)^{\infty} = 1 = L$$

That's not correct.

What conclusions do you get when

$$L = 1,\ L \neq 0$$

 

[Je m'appelle]

That's not correct.

You're absolutely correct, pardon my mistake, I must have been way too distracted.

To the original poster, disregard my last post, the correct evaluation would have been:

$$\lim_{n \rightarrow \infty} (1-\frac{1}{n})^n = L$$

Let us consider the following

$$y = (1-\frac{1}{n})^n$$

Now let's apply the natural log on both sides to get to the following

$$ln(y) = n \ ln(1-\frac{1}{n})$$

As $$n \rightarrow \infty$$ we have the following undefined expression

$$ln(y) = \infty . 0$$

At this point, we can use L'hôpital's at the right hand side of the equation, but first, we need to rewrite the expression in a fraction, in order to use L'hôpital's as follows

$$ln(y) = \frac{ln(1-\frac{1}{n})}{\frac{1}{n}}$$

Now we can use L'hôpital's

$$ln(y) = \frac{\frac{d}{dn}ln(1-\frac{1}{n})}{\frac{d}{dn}\frac{1}{n}}$$

To arrive at

$$ln(y) = -\frac{1}{(1-\frac{1}{n})}$$

Now we exponentiate both sides to get to

$$y = e^{-\frac{1}{(1-\frac{1}{n})}}$$

We can go back now to our original equation, substitute the new expression of y and evaluate the limit

$$\lim_{n \rightarrow \infty} y = L$$

$$\lim_{n \rightarrow \infty} e^{-\frac{1}{(1-\frac{1}{n})}} = L$$

$$e^{-1} = L, \ L = \frac{1}{e}$$

So as we found out that $$L \neq 0$$, the second condition for the Alternating Series Convergence Test is not satisfied and therefore the series diverges.