# Basic comparison test

1. Aug 13, 2009

### nameVoid

sum(1/n!,n,1,inf)
the only thing i can think of is
1/n^2>=1/n! , 1/n^2 being a convergent p series since p >1
thus /n! is convergent

Last edited: Aug 13, 2009
2. Aug 13, 2009

### Cyosis

nameVoid could you start putting some effort into your posts perhaps? In particular use the template and state your question plus your work clearly.

3. Aug 13, 2009

### nameVoid

whats the syntax for definite integral

4. Aug 13, 2009

### Cyosis

You should split off the first few terms to make the comparison with 1/n^2 more accurate.

The syntax for a definite integral is [ tex] \int_{lowerlimit}^{upperlimit} [ /tex].

This is not true for all n, take n=2 for example.

5. Aug 13, 2009

If you are investigating the series

$$\sum_{n=1}^\infty \frac{1}{n!}$$

then you can use

$$\sum_{n=1}^\infty \frac 1 {n^2}$$

and compare terms (and the second series IS a convergent p-series: I'm not sure what Cyosis was getting at).

6. Aug 13, 2009

### Cyosis

I was getting at the fact that for the first few terms the inequality doesn't hold. So concluding right away that one converges may bit a bit of a jump hence I suggested to write out the first few terms until the inequality indeed holds.

7. Aug 13, 2009

### nameVoid

$$\sum_{n=1}^{\infty}1/n!$$
$$\frac{1}{n!}\leq\frac{1}{n^2} , n\geq 4$$

$$\sum_{n=1}^{\infty}1/n^2:convergent$$
$$b_{n}=\sum_{n=4}^{\infty}1/n^2\geq\sum_{n=4}^{\infty}1/n!=a_{n}$$
i believe there is a property which states that if
$$\sum_{n=c}^{\infty}a_{n}$$
is convergent then the series converges for all n>=1 in any case here are the first few terms
$$\sum_{n=1}^{\infty}1/n^2=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}....$$
$$\sum_{n=1}^{\infty}1/n!=1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}....$$
this seems to confirm things

Last edited: Aug 13, 2009
8. Aug 13, 2009

### JG89

Looks fine to me. Good work. But you might want to prove that 1/n! <= 1/n^2 for n>= 4.

9. Aug 13, 2009

### Staff: Mentor

Just as a side note, a series almost identical to yours--
$$\sum_{n = 0}^{\infty}\frac{1}{n!}$$
-- converges to a number that is familiar to all mathematicians. Note that 0!, by definition, is 1.

10. Aug 13, 2009

### fmam3

Does the question state explicitly that you need to use the Comparison Test? If not, then you may want to try using the Ratio Test, which I think works equally well in this case and you don't need to do induction to check what you're comparing this against actually works.

Denote $$a_n = 1/n!$$. Then see that $$\lim_{n \to \infty} |a_{n+1}/a_n| = \lim \dfrac{1}{(n+1)!}\cdot n! = \lim \dfrac{n!}{(n+1)n!} = \lim \dfrac{1}{n+1} = 0 = \limsup |a_{n+1} / a_n| < 1$$. Thus, by the Ratio Test, this converges.