Thread: Proving that limit of sequence does not exists

1. Proving that limit of sequence does not exists

Hello,

I have some probelm regarding my homework.

I proved that the sequence $(-5)^{n}$ doesn't have limit and now I need to prove that the following sequence doesn't have limit too:
$\lim_{n \to \infty} \frac{(-5)^{n+1} + 5^n}{(-2)^n + 3}$
I tried to disprove it by the definition but it's doesn't go well so far, also I think it have to do something with the part that the sequence $(-5)^{n}$ doesn't have limit.

What is the best way to prove that this sequence doesn't have limit?

2. Re: Proving that limit of sequence does not exists

Originally Posted by kaze1
Hello,

I have some probelm regarding my homework.

I proved that the sequence $(-5)^{n}$ doesn't have limit and now I need to prove that the following sequence doesn't have limit too:
$\lim_{n \to \infty} \frac{(-5)^{n+1} + 5^n}{(-2)^n + 3}$
I tried to disprove it by the definition but it's doesn't go well so far, also I think it have to do something with the part that the sequence $(-5)^{n}$ doesn't have limit.

$\frac{(-5)^{n+1} + 5^n}{(-2)^n + 3}=\frac{(-1)^{n+1}(5)^{n+1} + 5^n}{(-1)^n(2)^n + 3}=\frac{(-1)^{n+1}(5) + 1}{(-1)^n\left(\tfrac{2}{5}\right)^n + 3\left(\frac{1}{5}\right)^n}$

Now what happens for n even or odd?

3. Re: Proving that limit of sequence does not exists

Jello, kaze1!

Another approach . . .

$\lim_{n \to \infty} \frac{(\text{-}5)^{n+1} + 5^n}{(\text{-}2)^n + 3}$

We have: . $\frac{\text{-}5(\text{-}5)^n + 5^n}{(\text{-}2)^n + 3} \;=\;\frac{\text{-}5(\text{-}1)^n(5^n) + 5^n}{(\text{-}1)^n(2^n) + 3} \;=\;\frac{5^n\big[1 - 5(\text{-}1)^n\big]}{3 + (\text{-}1)^n(2^n)}$

Divide numerator and denominator by $2^n:\;\;\frac{\left(\frac{5}{2}\right)^n\big[1 - 5(\text{-}1)^n\big]}{\frac{3}{2^n} + (\text{-}1)^n}$

Then: . $\lim_{n\to\infty}\frac{\left(\frac{5}{2}\right)^n \big[1 - 5(\text{-}1)^n\big]}{\frac{3}{2^n} + (\text{-}1)^n} \;=\;\frac{\infty\big[1-5(\text{-}1)^n\big]}{0 + (\text{-}1)^n}$

. . Even $n:\;\;\frac{\infty(1-5)}{1} \:=\:-\infty$

. . Odd $n: \;\;\frac{\infty(1+5)}{-1} \:=\:-\infty$

The limit diverges to $-\infty.$