# Proving limits

• Mar 4th 2010, 08:05 PM
Slazenger3
Proving limits
Prove that (5^n)/n! approaches 0 as n approaches infinity.
• Mar 4th 2010, 08:40 PM
Prove It
Quote:

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 $n! > 5^n$.
• Mar 4th 2010, 08:42 PM
Drexel28
Quote:

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, $\lim\text{ }\frac{n}{2n}\ne 0$
• Mar 4th 2010, 08:43 PM
Prove It
Quote:

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?
• Mar 4th 2010, 08:44 PM
Drexel28
Quote:

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.
• Mar 4th 2010, 08:45 PM
Prove It
Quote:

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 $n$.
• Mar 4th 2010, 08:48 PM
Drexel28
Quote:

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 $n!>5^n\implies \lim\text{ }\frac{5^n}{n!}=0$
• Mar 4th 2010, 10:05 PM
davismj
Quote:

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 $\displaystyle\lim_{n\to\infty} (a_n) = \displaystyle\lim_{n\to\infty}\bigg(\frac{a_{n+1}} {a_n}\bigg)$
• Mar 4th 2010, 11:07 PM
Prove It
Quote:

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.
• Mar 5th 2010, 04:00 AM
Drexel28
Quote:

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 $\frac{5^n}{n!}\to 0$
• Mar 5th 2010, 04:07 AM
Prove It
Quote:

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 (Clapping)