# Determine whethere the following series is convergent or divergent

## Homework Statement

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

## The Attempt at a Solution

I stopped.
I can not do anything.

## The Attempt at a Solution

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jgens
Gold Member
You do need to show us your attempt at the solution before we can help, so which convergence tests have you tried?

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 :/

Dick
Homework Helper
Write it as (n^n/(n+1)^n)*(1/(n+1)). The first factor approaches a finite limit. What is it?

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.

jgens
Gold Member
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 $n \geq 1$ can you find an upper bound for the following:

$$\frac{(n + 1)^{n+1}}{n^{n+1}}$$

Edit: Fixed typo

Edit 2: Fixed another typo. Sorry, I just seem to be mistake prone today.

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Dick
Homework Helper
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?

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 $$\sum_{n=1}^{\infty} \frac{1}{n+1}$$
The limit of the limit comparison test will give 1/e which is positive and finite number
hence, our series diverges.

Right?

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Dick
Homework Helper
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.