# Converging function

1. Sep 25, 2010

### ProPatto16

1. The problem statement, all variables and given/known data

determine whether $$\sum$$ n! / (n+2)! converges or diverges. if it converges, give the limit.

3. The attempt at a solution

i can see that this converges and the limit is 1/2

but i dunno how to mathematically show it... how can i factorise out the 1/2? so i have 1/2!*n!/n! which would give 1/2!*1=1/2

because n! / (n+2)! is not equal to n!/ (n!+2!)...

2. Sep 25, 2010

### JThompson

Do you mean

1. $$\sum\frac{n!}{(n+2)!}$$ or

2. $$\frac{\sum n!}{(n+2)!}$$

?

If #1, can't you simplify $$\frac{n!}{(n+2)!}=\frac{1}{(n+2)(n+1)}$$ ?

Last edited: Sep 25, 2010
3. Sep 25, 2010

### ProPatto16

the first one.

and i dunno, can you do that? ive never done any work with factorials. i had to google what they were!

4. Sep 25, 2010

### Staff: Mentor

Of course you can do that, based on how factorials are defined, usually something like this:
0! = 1
n! = n * (n - 1)!

So (n + 2)! = (n + 2) * (n + 1)! = (n + 2)(n + 1)*n!

5. Sep 25, 2010

### ProPatto16

" So (n + 2)! = (n + 2) * (n + 1)! = (n + 2)(n + 1)*n! "

so then 1 / (n + 2)(n + 1) * n!/n!
= 1 / (n+2)(n+1)

= 1 / (n2 + 3n +2)
now it looks like as n gets large, the limit is 0..

6. Sep 25, 2010

### Staff: Mentor

Yes, but that tells you exactly nothing. If you have a series $$\sum_{i = 1}^\infty}a_n$$ and $$\lim_{n \to \infty}a_n = 0$$, the series could converge or it could diverge.

Do you know any convergence tests you can use?

7. Sep 25, 2010

### ProPatto16

ratio and comparison testing would only prove convergence, without giving the limit.

integral test could work... let f(x) = 1/(n2+3n+2)

$$\int$$ f(x).dx with upper bound infinity (let that be a letter, say b) and lower bound 1. then take the definite integral and evaluate the limit as b gets large.

8. Sep 25, 2010

### ProPatto16

and that integral is = log(x+1) - log(x+2)
= log[(x+1)/(x+2)]

am i goin the right direction? i dunno where to go from there..

9. Sep 26, 2010

### Staff: Mentor

The problem first asks you whether the series converges or diverges. Have you come up with a good reason for thinking that this series converges. "I can see that it converges" is not a good reason.

The integral test tells you only that a series converges or diverges. It doesn't give you the sum of a convergent series.

Hint: partial fractions...

10. Sep 26, 2010

### ProPatto16

use the limit comparison test, with bn = 1/n2... this will show it converges.

11. Sep 26, 2010

### ProPatto16

but then im still not sure how to evaluate the limit to show the limit of convergence.... if i multiply through by 1/n i end up with 0 as the numerator as n gets large, so that doesnt work. i dont know how else to go about it. bit dense when it comes to these.

12. Sep 26, 2010

### fzero

Follow through on Mark44's hint about partial fractions. Use partial fractions to write the partial sums as a difference between two terms. You will find some cancellations.

13. Sep 26, 2010

### ProPatto16

like this...

1/((n+1)(n+2))
by decompostion becomes 1/(n+1) - 1/(n+2)

so (1/2-1/3)+(1/3-1/4)+(1/4-1/5)+....(1/(n+1)-1/(n+2))

= 1/2 - 1/(n+2)

as n gets large, 1/(n+2) becomes 0 so:

= 1/2-0
=1/2

there.

14. Sep 26, 2010

### Staff: Mentor

That's the basic idea, yes. Here's what the cleaned-up version looks like, using partial sums.

$$S_n = \sum_{k = 1}^n \frac{1}{(k + 1)(k + 2)} = \sum_{k = 1}^n \left(\frac{1}{k + 1} - \frac{1}{k + 2} \right) = \frac{1}{2} - \frac{1}{n + 2}$$
$$\lim_{n \to \infty} S_n = \frac{1}{2}$$

15. Sep 26, 2010

### ProPatto16

yeahh im not the best at using the functions and stuff on here haha. thanks for the help though!