# Thread: Limiit of a sequence

1. ## Limiit of a sequence

For the following sequence : ((-1)^n) / (n!), if the sequence converges find its limit.

How can you mathematically show that the limit is 0, i can see from observation that as n increases the denominator gets bigger and the term gets closer to 0, but how can i show it mathematically.

This is actually practice question homework 4 marks so i'm a bit apprehensive about just saying that the limit is 0 from observation without any mathematical proof.

Also by applying d'alemberts ratio test i get 0, is the result of the d'alembert ratio test the limit?

2. ## Re: Limiit of a sequence

$\displaystyle \dfrac{-1}{n!} \le \dfrac{(-1)^n}{n!} \le \dfrac{1}{n!}$

By the Squeeze Theorem, since $\displaystyle \lim_{n \to \infty} \dfrac{-1}{n!} = \lim_{n\to \infty} \dfrac{1}{n!} = 0$, you have $\displaystyle \lim_{n \to \infty} \dfrac{(-1)^n}{n!} = 0$.

To show that $\displaystyle \lim_{n \to \infty} \dfrac{-1}{n!} = 0 = \lim_{n \to \infty} \dfrac{1}{n!}$, you can use an $\displaystyle \varepsilon$ proof. Given $\displaystyle \varepsilon>0$, can you show that there exists $\displaystyle N \in \Bbb{N}$ such that for all $\displaystyle n\ge N$, $\displaystyle \left|\dfrac{-1}{n!} - 0 \right| < \varepsilon$ and $\displaystyle \left|\dfrac{1}{n!} - 0 \right| < \varepsilon$? First, let's use an induction argument to show that for all $\displaystyle n\in \Bbb{N}, n!\ge 2^{n-1}$.

Base of induction: $\displaystyle n=0$. Then $\displaystyle 0! = 1 \ge 2^{0-1} = \dfrac{1}{2}$ is true. $\displaystyle n=1$. Then $\displaystyle 1! = 1 \ge 2^{1-1} = 1$ is also true.

Assume the proposition is true for some positive integer $\displaystyle n$ (we already showed it is true for $\displaystyle n=1$, so this assumption is true). Then $\displaystyle (n+1)! = (n+1)n!$. For all $\displaystyle n\ge 1$, $\displaystyle n+1\ge 2$. So, $\displaystyle (n+1)n! \ge 2n! \ge 2\cdot 2^{n-1} = 2^{(n+1)-1}$ (the second inequality came from application of the induction assumption). Hence, $\displaystyle n! \ge 2^{n-1}$ for all $\displaystyle n$.

Now that is proven, choose any positive integer $\displaystyle N > 1-\log_2 \varepsilon$. Then for any $\displaystyle n\ge N$, we have $\displaystyle n > 1-\log_2 \varepsilon$, which implies $\displaystyle 2^n > \dfrac{2}{\varepsilon}$. Dividing both sides by 2 gives $\displaystyle n! \ge 2^{n-1}>\dfrac{1}{\varepsilon}$. Taking the reciprocal of both sides flips the inequality: $\displaystyle \dfrac{1}{n!} < \varepsilon$. This is exactly what you wanted to prove.

There are various other ways of proving it.

3. ## Re: Limiit of a sequence

Thank you very much, your explanation was very good and helped me learn a lot. however for this question is there a simple way to do this as i have very limited space provided to do this question so i think that my teacher wanted me to do it very quickly in a simple way, as its only worth 3 marks in a 100 mark paper. So is there a shorter way of mathematically proving limit is 0?

4. ## Re: Limiit of a sequence

Given $\displaystyle \varepsilon>0$, choose any positive integer $\displaystyle N>1-\log_2 \varepsilon$. For any $\displaystyle n\ge N$, we have:

$\displaystyle \left|\dfrac{(-1)^n}{n!}-0\right| = \dfrac{1}{n!} \le \dfrac{1}{2^{n-1}} \le \dfrac{1}{2^{N-1}} < \dfrac{1}{2^{1-\log_2 \varepsilon-1}} = \varepsilon$. That's really all you need. If there is a question about the fact that $\displaystyle 2^{n-1} \le n!$, your instructor can ask you to prove it, but it is obviously true as $\displaystyle 2^{n-1} = \underbrace{2\cdots 2}_{n-1\text{ times}}$ while $\displaystyle n! = \underbrace{n\cdot (n-1)\cdots 2\cdot 1}_{n\text{ terms}}$ is a product of $\displaystyle n$ numbers, all but one of them are at least 2.

5. ## Re: Limiit of a sequence

An even easier argument. $\displaystyle n! \ge n$ for all $\displaystyle n>0$

6. ## Re: Limiit of a sequence

Ok, thank you very much for your help.