How can understanding limits help solve infinite series in Calculus II?

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Homework Help Overview

The discussion revolves around the evaluation of the infinite series \(\sum_{n=1}^\infty(-1)^n \frac{n^n}{n!}\) and the application of limits in this context, specifically as \(n\) approaches infinity.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the limit \(\lim_{n \rightarrow \infty} \frac{n^n}{n!}\) and question whether it approaches zero or another value. There are attempts to analyze the behavior of the series and its components, with some participants expressing confusion over the application of limits and the concept of indeterminate forms.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have offered insights into the nature of the limit and the behavior of the series, while others express difficulty in grasping the concepts involved. There is a recognition of the need to revisit foundational ideas related to limits and infinite series.

Contextual Notes

Participants note a lack of understanding of the underlying concepts of calculus, particularly in relation to limits and series, which may be affecting their ability to engage with the problem effectively.

MillerGenuine
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Homework Statement



[tex] \sum_{n=1}^\infty(-1)^n \frac{n^n}{n!}[/tex]

Homework Equations



I can not find my limit as n approaches infinity. I know that the answer is infinity but I am not sure how to get it.


The Attempt at a Solution

 
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n^n = n*n*n*n*n*...*n ( ntimes)
n! = 1 *2*3 ... *n

What can you say about

[tex] <br /> \lim_{n \rightarrow \infty} \frac{n^n}{n!}<br /> [/tex]

Does it go to zero ?

EDIT
Fixed
 
Seems to me that it would become [tex]\frac{\infty}{infty}=1[/tex]
 
MillerGenuine said:
Seems to me that it would become [tex]\frac{\infty}{infty}=1[/tex]

How did you arrive at that answer ?

It is not correct.

[tex]\lim_{n \rightarrow \infty} \frac{n^n}{n!} = \lim_{n \rightarrow \infty} \frac{n*n*n*...*n}{1*2*3...*n} = \lim_{n \rightarrow \infty} \frac{n*n*n*...*n}{1*2*3...*n-1}<br /> <br /> [/tex]
As n gets very large

n/1 = n
n/2 > 1
n/3 > 1
...
n/n-1 [tex]\cong 1[/tex]
 
Last edited:
I just based it on the fact that i have an "n" in the numerator and an "n" in the denominator so plugging in infinity for both will give me 1. Which is clearly incorrect, but i just have no idea how to approach this problem.
 
MillerGenuine said:
I just based it on the fact that i have an "n" in the numerator and an "n" in the denominator so plugging in infinity for both will give me 1. Which is clearly incorrect, but i just have no idea how to approach this problem.

Well look at what I gave you .

Or simply look at a pattern

Lets use some examples

n=1

1^1/ 1! =1

2^2 /2! = 4/2 =2

3^3 /3! = 9/2



Can you see that n^n /n! is increasing and not bounded above.

In fact

n^n/n! > = n
 
Im sorry but its just not clicking for me. I see what you have shown and all i keep seeing in my head is infinity^infinity/infinity! which seems to be infinity/infinity=1
my apoogies if this sounds stupid but bare with me
 
MillerGenuine said:
Im sorry but its just not clicking for me. I see what you have shown and all i keep seeing in my head is infinity^infinity/infinity! which seems to be infinity/infinity=1
my apoogies if this sounds stupid but bare with me
Don't plug in infinity like that. Your thinking is dangerous :-p

If we followed your logic then we could prove that
[tex]\lim_{n \rightarrow \infty} \frac{n}{n^2 +1} = 1[/tex]

Read what I posted carefully maybe something would click or go to bed and attempt the problem tomorrow ( assuming it's night where you are it's 1:27 am here :wink:).
 
MillerGenuine said:
Seems to me that it would become [tex]\frac{\infty}{\infty}=1[/tex]
[tex]\frac{\infty}{\infty}[/tex] is not a number. This is one of several indeterminate forms, others of which include 0/0, [tex]\infty - \infty, and[/tex], [tex]1^{\infty}[/tex]

These are all symbolic and do not represent numbers. When you have a limit expression whose limit is any of these, the actual limit can come out to be anything.

Going back to what ╔(σ_σ)╝ said, an for your series is nn/n!. This is
[tex]\frac{n \cdot n \cdot n \cdot ... \cdot n}{1 \cdot 2 \cdot 3 \cdot ... \cdot n}[/tex]

or
[tex]\frac{n}{1} \cdot \frac{n}{2} \cdot \frac{n}{3} \cdot ... \cdot \frac{n}{n}[/tex]

As n gets large, the last factor on the right remains 1, but what is happening to the factors on the left?
 
  • #10
Read what I posted carefully maybe something would click or go to bed and attempt the problem tomorrow
I think i may take your advice on going to bed & take a fresh look at it tomorrow because i just can not seem to understand this. I am sure if either of you were to explain in person It would be much easier to understand. I think the main problem is my lack of understanding the concepts of this class. my prof only teaches mechanics of problems so i struggle with conceptual calculus
 
  • #11
MillerGenuine said:
I think i may take your advice on going to bed & take a fresh look at it tomorrow because i just can not seem to understand this. I am sure if either of you were to explain in person It would be much easier to understand. I think the main problem is my lack of understanding the concepts of this class. my prof only teaches mechanics of problems so i struggle with conceptual calculus

It's okay; my brain also I not functioning as I would like at this point too :).

Take some time to review limits as this is indispensable to discussions on infinite series.
 

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