1. ## Proving limits

Prove that (5^n)/n! approaches 0 as n approaches infinity.

2. Originally Posted by Slazenger3
Prove that (5^n)/n! approaches 0 as n approaches infinity.
All you need to prove is that for large values of $n$ that $n! > 5^n$.

3. Originally Posted by Prove It
All you need to prove is that for large values of $n$ that $n! > 5^n$.
For large values of $n$ we see that $2n>n$. But, $\lim\text{ }\frac{n}{2n}\ne 0$

4. Originally Posted by Drexel28
For large values of $n$ we see that $2n>n$. But, $\lim\text{ }\frac{n}{2n}\ne 0$
Yes but we're not proving it for $\frac{n}{2n}$ are we?

5. Originally Posted by Prove It
Yes but we're not proving it for $\frac{n}{2n}$ are we?
Yes. But proving that one is larger than the other is not sufficient to show that the limit is zero.

6. Originally Posted by Drexel28
Yes. But proving that one is larger than the other is not sufficient to show that the limit is zero.
In this case it would be.

I think it is bleedingly obvious that

$\frac{5 \cdot 5 \cdot 5 \dots 5}{n(n - 1)(n - 2)\dots 3 \cdot 2}$

must go to $0$ since the denominator is so much bigger than the numerator for large values of $n$.

7. Originally Posted by Prove It
In this case it would be.

I think it is bleedingly obvious that

$\frac{5 \cdot 5 \cdot 5 \dots 5}{n(n - 1)(n - 2)\dots 3 \cdot 2}$

must go to $0$ since the denominator is so much bigger than the numerator for large values of $n$.
Bleedingly obvious as this case may be. That does not prove the assertion that proving $n!>5^n\implies \lim\text{ }\frac{5^n}{n!}=0$

8. Originally Posted by Drexel28
Bleedingly obvious as this case may be. That does not prove the assertion that proving $n!>5^n\implies \lim\text{ }\frac{5^n}{n!}=0$
A simple application of the ratio test shows that $\displaystyle\lim_{n\to\infty}\frac{5^n}{n!} = \displaystyle\lim_{n\to\infty}\frac{5^{n+1}n!}{5^n (n+1)!} = \displaystyle\lim_{n\to\infty}\frac{5}{n} = 0$

For reference, the ratio test says that $\displaystyle\lim_{n\to\infty} (a_n) = \displaystyle\lim_{n\to\infty}\bigg(\frac{a_{n+1}} {a_n}\bigg)$

9. Originally Posted by davismj
A simple application of the ratio test shows that $\displaystyle\lim_{n\to\infty}\frac{5^n}{n!} = \displaystyle\lim_{n\to\infty}\frac{5^{n+1}n!}{5^n (n+1)!} = \displaystyle\lim_{n\to\infty}\frac{5}{n} = 0$

For reference, the ratio test says that $\displaystyle\lim_{n\to\infty} (a_n) = \displaystyle\lim_{n\to\infty}\bigg(\frac{a_{n+1}} {a_n}\bigg)$
This is not a series. The ratio test only determines the convergence or divergence of a series. Not only that, it does not evaluate the series, it only tells us if the series is convergent or divergent.

10. Originally Posted by Prove It
This is not a series. The ratio test only determines the convergence or divergence of a series. Not only that, it does not evaluate the series, it only tells us if the series is convergent or divergent.
Yes. But he proved that $\sum_{n=1}^{\infty}\frac{5^n}{n!}$ converges, and thus by necessity $\frac{5^n}{n!}\to 0$

11. Originally Posted by Drexel28
Yes. But he proved that $\sum_{n=1}^{\infty}\frac{5^n}{n!}$ converges, and thus by necessity $\frac{5^n}{n!}\to 0$
Good point