Converging Function Homework: Determine Limit (1/2)

[ProPatto16]

Homework Statement

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

The Attempt at a Solution

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

but i don't know 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!)...

 

[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)} ?

 

[ProPatto16]

the first one.

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

 

[Mark44]

the first one.

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

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!

 

[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..

 

[Mark44]

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?

 

[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.

 

[ProPatto16]

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

am i goin the right direction? i don't know where to go from there..

 

[Mark44]

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...

 

[ProPatto16]

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

 

[ProPatto16]

but then I am 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 doesn't work. i don't know how else to go about it. bit dense when it comes to these.

 

[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.

 

[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.

 

[Mark44]

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}

 

[ProPatto16]

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