Question

Evaluate the limit if it exists.

Answer 1

\[\lim_{n \rightarrow \infty}\frac{ (-1)^n }{ n }\]

Answer 2

\[\lim_{n \rightarrow \infty} \frac{C}{n} = 0\]
where C is either 1 or -1

Answer 3

The numerator is not a constant 1 or -1 though...
it's 1 and -1 back and forth over and over, alternating.
I agree that it zeroes out, but not for the reason you stated.

Is there some type of test we can do to show this?
I can't remember >.< Grrr

Answer 4

\[\lim_{n \rightarrow \infty}\left| \frac{ (-1)^n }{ n } \right|=\lim_{n \rightarrow \infty}\frac{ 1 }{ n }=0\]
Using Theorem 6: If \[\lim_{n \rightarrow \infty}\left| a _{n} \right|=0\],then \[\lim_{n \rightarrow \infty}a _{n}=0.\]
Just copying the example from the book, but I still don't understand it. Hehe

Answer 5

Having trouble understanding it? :o
For all of the terms that have a -1 in the numerator, it terms them into a +1.

Answer 6

I don't get how \[\lim_{n \rightarrow \infty}\left| \frac{ (-1)^n }{ n } \right|\] becomes \[\lim_{n \rightarrow \infty}\frac{ 1 }{ n }\]

Answer 7

For odd n, \(\large\rm |(-1)^n|=|-(1)^n|=|-1|=1\).
We can pull the negative outside when the power is odd, ya?

For even n, we don't even need to bother with the absolute value,
the even power squares it away,
\(\large\rm |(-1)^n|=|(1)|=1\)

Answer 8

Why does the denominator left unchanged?

Answer 9

We're plugging in `positive values for n`.
So for every n, \(\large\rm |n|=n\).
They remain unchanged, ya?

Answer 10

|73|=73

Answer 11

From my understanding, the numerator fluctuates between 1 and -1 (after solving for the numerator, so the n exponent disappears), but since there is an absolute value, it will always be positive 1.

Answer 12

Okay, I think I understand it now.

Answer 13

Thanks all :)

Answer 14

cool c: