# Converging Function Homework: Determine Limit (1/2)

• ProPatto16
In summary, the homework statement is to determine whether the sum of n! / (n+2)! converges or diverges. If it converges, give the limit. However, the problem asks for a good reason for thinking that this series converges, and the student does not provide a good reason. The integral test tells the student only that the series converges or diverges, but does not give the sum of a convergent series. Partial fractions can be used to help with the partial sums of the series, and the limit comparison test can be used to show that the series converges.
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!)...

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:
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!

ProPatto16 said:
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!

" 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..

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?

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.

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..

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...

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

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.

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.

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.

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

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

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

## 1. What is the purpose of determining a limit in converging function homework?

The purpose of determining a limit in converging function homework is to understand the behavior of a function as it approaches a specific value. It helps in determining the behavior of a function at a given point, even if the function is undefined at that point.

## 2. How do you determine a limit of a converging function?

To determine the limit of a converging function, you need to evaluate the function at values close to the given point and see if the function approaches a single value or if it becomes undefined. This process is known as direct substitution. You can also use algebraic manipulation or graphing techniques to determine the limit.

## 3. What is the difference between a left-hand limit and a right-hand limit?

A left-hand limit is the value that a function approaches from the left side of a specific point, while a right-hand limit is the value that a function approaches from the right side of a specific point. The left-hand and right-hand limits may not always be equal, and the limit only exists if both the left-hand and right-hand limits are equal.

## 4. Can a limit of a function be undefined?

Yes, a limit of a function can be undefined. This happens when the left-hand and right-hand limits do not approach the same value, or when the function is discontinuous at the given point. In such cases, the limit does not exist.

## 5. How do you determine if a limit exists?

A limit exists if the left-hand and right-hand limits approach the same value, or if the function is continuous at the given point. You can also use the formal definition of a limit to determine if it exists by checking if the function approaches a specific value as the input approaches a given value.

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