Determine whethere the following series is convergent or divergent

  • #1

Homework Statement


Determine whether the following series is convergent or divergent:
[tex]
\frac{1}{2^2}+\frac{2^2}{3^3}+\frac{3^3}{4^4}+....
[/tex]
I rewrite it as:
[tex]
\sum_{n=1}^{\infty} \frac{n^n}{(n+1)^{n+1}}
[/tex]

Homework Equations





The Attempt at a Solution


I stopped.
I can not do anything.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
jgens
Gold Member
1,581
50
You do need to show us your attempt at the solution before we can help, so which convergence tests have you tried?
 
  • #3
You do need to show us your attempt at the solution before we can help, so which convergence tests have you tried?
Its not geometric
not telescoping
nth term test failed
can not use integral test
Ratio & Root failed because its algebric function

There is only the comparison tests
I tried them a lot
but every time i failed :/
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
619
Write it as (n^n/(n+1)^n)*(1/(n+1)). The first factor approaches a finite limit. What is it?
 
  • #5
Write it as (n^n/(n+1)^n)*(1/(n+1)). The first factor approaches a finite limit. What is it?
The first one goes to 1/e
the second goes to 0
1/e times 0 = 0


nth term test failed.
 
  • #6
jgens
Gold Member
1,581
50
Alright, I would suggest using the comparison test for this particular series (although I'm sure that there's probably a simpler way to do it). If [itex]n \geq 1[/itex] can you find an upper bound for the following:

[tex]\frac{(n + 1)^{n+1}}{n^{n+1}}[/tex]

Edit: Fixed typo

Edit 2: Fixed another typo. Sorry, I just seem to be mistake prone today.
 
Last edited:
  • #7
Dick
Science Advisor
Homework Helper
26,258
619
The first one goes to 1/e
the second goes to 0
1/e times 0 = 0


nth term test failed.
What 'nth term test' are you talking about? You are right the first factor approaches 1/e. So for large values of n a term in the series is approximately 1/(e*(n+1)). What kind of test does that suggest?
 
  • #8
What 'nth term test' are you talking about? You are right the first factor approaches 1/e. So for large values of n a term in the series is approximately 1/(e*(n+1)). What kind of test does that suggest?
nth terms test = Test for divergence.

hmmm .. I got it .. Limit comparison test with the divergent series [tex]\sum_{n=1}^{\infty} \frac{1}{n+1}[/tex]
The limit of the limit comparison test will give 1/e which is positive and finite number
hence, our series diverges.

Right?
 
Last edited:
  • #9
Dick
Science Advisor
Homework Helper
26,258
619
nth terms test = Test for divergence.

hmmm .. I got it .. Limit comparison test with the divergent series [tex]\frac{1}{n+1}[\tex]
The limit of the limit comparison test will give 1/e which is positive and finite number
hence, our series diverges.

Right?
Right.
 

Related Threads on Determine whethere the following series is convergent or divergent

Replies
3
Views
340
Replies
2
Views
2K
Replies
2
Views
582
Replies
1
Views
3K
Replies
11
Views
9K
Replies
15
Views
6K
Replies
2
Views
5K
Replies
1
Views
1K
Replies
3
Views
4K
Replies
2
Views
2K
Top