# Thread: Using squeeze theorem to determine if a sequence is convergent or divergent

1. ## Using squeeze theorem to determine if a sequence is convergent or divergent

The question states to determine whether or not the sequence is convergent or divergent, in class we used squeeze theorem on a similar example to solve this question. Can someone help me do the same with this question? I'm having a particular amount of trouble selecting the bounds.

It also asks if it is convergent, to find its limit. Any help would be greatly appreciated, thanks!!

The sequence is {((-10)^n)/n!}

2. ## Re: Using squeeze theorem to determine if a sequence is convergent or divergent

For $n\ge 80$, it is easy to show that $n!>20^n$. So $\dfrac{(-10)^n}{n!}<\dfrac{(-10)^n}{20^n}$. Also, $0 < \dfrac{(-10)^n}{n!}$. So, this gives you two sequences to squeeze. Both have the same limit: 0.

Proof that $n!>20^n$ for $n\ge 80$. At $n=80$, we have $80! \approx 1.56\times 10^{118}$ while $20^{80} \approx 1.21\times 10^{104}$. So, the claim is true for $n=80$. Assume the claim is true for some positive integer $k\ge 80$. Then $(k+1)! = (k+1)k!>(k+1)20^k\ge 81\cdot 20^k > 20\cdot 20^k = 20^{k+1}$. By induction, the claim is true for all $n\ge 80$.

Let $a_n = 0$, $b_n = \dfrac{(-10)^n}{n!}$, $c_n = \left(-\dfrac{1}{2}\right)^n$

For all $n\ge 80$, we have $a_n < b_n < c_n$ and $\displaystyle \lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = 0$. Therefore, by the Squeeze Theorem, you have $\displaystyle \lim_{n\to \infty} b_n = 0$.

3. ## Re: Using squeeze theorem to determine if a sequence is convergent or divergent

Originally Posted by SlipEternal
For $n\ge 80$, it is easy to show that $n!>20^n$. So $\color{red}{\dfrac{(-10)^n}{n!}<\dfrac{(-10)^n}{20^n}}$. Also, $0 < \dfrac{(-10)^n}{n!}$. So, this gives you two sequences to squeeze. Both have the same limit: 0.
Did you note that $(-10)^n$ is alternating? In fact $\dfrac{(-10)^{81}}{81!}>\dfrac{(-10)^{81}}{20^{81}}$.

But easily corrected I think.

4. ## Re: Using squeeze theorem to determine if a sequence is convergent or divergent

Originally Posted by Plato
Did you note that $(-10)^n$ is alternating? In fact $\dfrac{(-10)^{81}}{81!}>\dfrac{(-10)^{81}}{20^{81}}$.

But easily corrected I think.
Oh, right. Lol.

$a_n = -\left(\dfrac{1}{2}\right)^n$

$c_n = \left(\dfrac{1}{2}\right)^n$

Now, for all $n\ge 80$ we have $a_n <b_n <c_n$.

5. ## Re: Using squeeze theorem to determine if a sequence is convergent or divergent

Originally Posted by SlipEternal
$a_n = -\left(\dfrac{1}{2}\right)^n$
$c_n = \left(\dfrac{1}{2}\right)^n$
Are you really serious about any of this? Your replies say otherwise.

Do you mean $-\left(\dfrac{1}{2}\right)^n\le\left(\dfrac{-1}{2}\right)^n\le\left(\dfrac{1}{2}\right)^n$

If not then what is your point? Do you understand any of this?

6. ## Re: Using squeeze theorem to determine if a sequence is convergent or divergent

Originally Posted by Plato
Are you really serious about any of this? Your replies say otherwise.

Do you mean $-\left(\dfrac{1}{2}\right)^n\le\left(\dfrac{-1}{2}\right)^n\le\left(\dfrac{1}{2}\right)^n$

If not then what is your point? Do you understand any of this?
Yes, I am serious. My point was that by changing $a_n$ to negative $c_n$, I bound the alternating sequence $b_n$ above and below allowing the Squeeze Theorem to apply.

What part of that do you not understand? I'll try to break it down more for you.