Compare & Determine: The 1/n! Series Convergence/Divergence

[belvol16]

Homework Statement

Determine whether the series converges or diverges.
∞
∑ 1/n!
n=1

Homework Equations

If ∑b_n is convergent and a_n≤b_n for all n, then ∑a_n is also convergent.
Suppose that ∑a_n and ∑b_n are series with positive terms. If
lim a_n = C
n→∞ b_n
where c is finite number and c>o, then either both series converge or both diverge.

The Attempt at a Solution

So I said:
1/n! < 1/n . And 1/n diverges because it is a harmonic series.
Then I tried the limit comparison test...
lim (1/n!)/(1/n) = ∞/o.
n→∞
I want to use the L'Hospital rule...but I have no clue how to derive that.
The book says the answer is convergent...I'm not sure if I'm missing something.

Thanks for your help!

 

[fresh_42]

Do you know any convergent series, yet? Or what other criteria beside the direct comparison test do you know already?

 

[belvol16]

The book has the problem listed under the comparison tests.
But we have also gone over The Integral test, Alternating Series, and Absolute Convergence and the Ratio and Root Tests.

 

[fresh_42]

The comparison test works well, if you know other convergent series. Also for the limit comparison you need to know another convergent series.
How about the ratio test, if you don't have series for comparison at hand?

 

[belvol16]

So I just used the ratio test and found that:
lim |(1/(n+1)!)/(1/n)| = 0 <1 therefore it is absolutely convergent.
n→∞
However, I am still wondering how one would solve this with the comparison test.
I know that 1/n! < 1/n² and 1/n² is convergent because p=2>1.
Would this be the way to solve it with the comparison test? If so I'm confused how one would be able to compare 1/n! to 1/n² because we don't have 1/(n!)²

 

[fresh_42]

If you know that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, then this is fine and you are correct with what you said.
You don't need $\frac{1}{(n!)^2}$. Comparison means term by term: $a_n < b_n$ and in your case $\frac{1}{n!} < \frac{1}{n^2}$ for all $n > 3$. And finitely many terms (here the first three) don't affect convergence.

 

[Mark44]

So I just used the ratio test and found that:
lim |(1/(n+1)!)/(1/n)| = 0 <1 therefore it is absolutely convergent.
n→∞

All you need to say is that the series is convergent. It's true that it is absolutely convergent, but trivially so. This term is usually applied to series whose terms include negative numbers. Note that $\sum_{i = 1}^{\infty}\frac{(-1)^n}{n}$ is convergent, but not absolutely convergent. (The series itself converges, but the series made up of the absolute values of this series, diverges.

 

[belvol16]

If you know that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, then this is fine and you are correct with what you said.
You don't need $\frac{1}{(n!)^2}$. Comparison means term by term: $a_n < b_n$ and in your case $\frac{1}{n!} < \frac{1}{n^2}$ for all $n > 3$. And finitely many terms (here the first three) don't affect convergence.

I guess I don't understand why the comparison test is applicable if you could obtain two different results. In one case it is 1/n!<1/n and that is divergent, but in another it is 1/n!<1/n^2 is convergent. I understand both of these inequalities are true and that 1/n! < 1/n^p is convergent for all p>1...but how is one able to just look past 1/n ?

 

[fresh_42]

You can always have a divergent series which is greater, whether it is $\sum_{n=1}^\infty \frac{1}{n}$ or simply $\sum_{n=1}^\infty 1$.
But how can it be arbitrary large, if it is bounded from above by a convergent sum? We have $0 < \sum_{n=1}^\infty a_n < \sum_{n=1}^\infty b_n < \sum_{n=1}^\infty c_n$. If $\sum_{n=1}^\infty a_n$ is trapped between $0$ and $\sum_{n=1}^\infty b_n = B < \infty \,$, how can it "explode"? The sum $\sum_{n=1}^\infty c_n$ simply doesn't play any role. In the case above it is
$$0 < \sum_{n=1}^\infty \frac{1}{n!} = e-1 \approx 1.72 < 3+\sum_{n=1}^\infty \frac{1}{n^2} = 3+\frac{\pi^2}{6} \approx 4.64 < \sum_{n=1}^\infty \frac{1}{n} = \infty$$
(The $3$ shows up, because the first three terms $\frac{1}{n^2}$ aren't greater than $\frac{1}{n!}$ and $3$ is an estimation without thinking about it.)

 

[Mark44]

I guess I don't understand why the comparison test is applicable if you could obtain two different results.

You aren't getting two different results.

In one case it is 1/n!<1/n and that is divergent, but in another it is 1/n!<1/n^2 is convergent. I understand both of these inequalities are true and that 1/n! < 1/n^p is convergent for all p>1...but how is one able to just look past 1/n ?

Suppose that $\sum c_n$ is a known convergent series, and that $\sum d_n$ is a known divergent series. (c and d are chosen to be suggestive of convergence and divergence.)
Suppose also that you need to investigate the convergence or divergence of $\sum a_n$.

There are four possible comparisons you could do with the two series whose behavior is known.
1) $a_n > c_n$
2) $a_n < c_n$
3) $a_n > d_n$
4) $a_n < d_n$

Only two of the four inequalities are helpful: (2) and (3).

For (1), determining that $a_n$ is larger than the corresponding term of a convergent series does us no good at all. Likewise, determining that $a_n$ is smaller than the corresponding term of a divergent series doesn't necessarily imply that $\sum a_n$ converges.

The upshot of all of this is that when you use the Comparison Test, you need to have an idea about whether the series you're working with converges or diverges. If you think your series converges, you need to show that the terms of this series are smaller than those of some convergent series. If you think your series diverges, you need to show that the terms of this series are larger than those of some divergent series.