# Thread: finding the limit of the sequence

1. ## finding the limit of the sequence

finding this limit:

lim
n -> infinity (e^-n(-3)^n)

After trying to graph this, i just get errors.
I assume its because i tried to raise a natural log to a negative number?

i imagine that this does not have a limit?
how can i prove this mathematically.

thankyou.

2. Originally Posted by rcmango
finding this limit:

lim
n -> infinity (e^-n(-3)^n)

After trying to graph this, i just get errors.
I assume its because i tried to raise a natural log to a negative number?

i imagine that this does not have a limit?
how can i prove this mathematically.

thankyou.
I assume you mean:

$\lim_{n \to \infty} e^{-n}\,(-3)^n$

Rearrange this:

$\lim_{n \to \infty} e^{-n}\,(-3)^n=\lim_{n \to \infty} (-1)^n e^{-n}\,(3)^n =$ $\lim_{n \to \infty}(-1)^n e^{-n}\,e^{n\,\ln(3)} =\lim_{n \to \infty}(-1)^n e^{-n+n\,\ln(3)}$

..... $=\lim_{n \to \infty}(-1)^n e^{(\ln(3)-1)n}$

Now the coefficient of $n$ in the exponential is $>0$ so the exponential term goes to infinity, while the powers of $-1$ change the sign of alternate terms. So the absolute value of this sequence goes to infinity, but with indeterminate sign, that is there is no limit.

RonL

3. very nice, Thankyou.