# Homework Help: Series- convergent or divergent?

1. Feb 2, 2008

### nuclearrape66

from n=1 to infinity

does the series converge or diverge?

n!/n^n

its in the secition of the book with the comparison test and limit comparison test.

if you compare it with 1/n^n (this is a geomoetric series) you get a= 1/n amd r= 1/n

but in the thrm r = to some finite number and i dont know if the geometric series thrm applies here.

2. Feb 2, 2008

### StatusX

A geometric series has a constant factor between successive elements, so 1/n^n does not qualify. Look at:

$$\frac{n!/n^n}{(n-1)!/(n-1)^{n-1}}$$

3. Feb 2, 2008

### nuclearrape66

statusx....help me solve my problem rather than leading me to different problems

if geometric series does'nt apply here, what shouudl be done when comparing

4. Feb 2, 2008

### quasar987

Re-read StatusX's post. The last part of his message is a major hint.

5. Feb 2, 2008

### nuclearrape66

hmmm what does (n-1)! = ?

n!(n-1)?

6. Feb 2, 2008

### Mystic998

It means take n, subtract 1, then apply the factorial operator to the result.

7. Feb 2, 2008

### HallsofIvy

No, (n+1)!= n!(n+1) which is what you may be thinking of. It should be obvious that (n-1)! is less than n!.

8. Feb 2, 2008

### foxjwill

I was able to solve this using the limit test and some strategic use of logarithms. Basically, set the limit of the limit test equal to $$y$$,

$$y &= \lim_{n \rightarrow \infty} \left | \left ( \frac{n}{n+1} \right )^n \right |\\$$

take the ln of both sides,

$$\ln y &= \lim_{n \rightarrow \infty} n {\ln \left |\frac{n}{n+1}\right | }$$

then use L'Hopitals Rule, and solve for $$y$$.

9. Feb 2, 2008

### nuclearrape66

can anyone write the solution?

please dont whore my thread. i just need the solution to the problem, i'm not lookign for hints.

thanks

10. Feb 2, 2008

### foxjwill

Unfortunately, unless you show us you're work (i.e. that you've tried what we've suggested), we can't write the whole solution. However, I can tell that yes, it does converge. Oh, and my suggestion used the ratio test, not the limit test. >_<

11. Feb 2, 2008

### nuclearrape66

oh okay...ratio test. now i see

12. Feb 2, 2008

### Gib Z

Why do you think you needed to give advice to a homework helper? They got the badge because they have a somewhat decent knowledge of the problems they give advice on, and know how to help. It they lead you to a different problem, then theres a good chance it will also solve the original one.

Help is not "whoring a thread". If you just want a solution, do it yourself and quit wasting our time.

PS. Is there any reason you chose your username to be NuclearRape66?

13. Feb 2, 2008

### nuclearrape66

are you gonna go through my inbox?

and what you're doing right there is whoring my thread...if you dont wanan help dont help. dotn speak for everyone, speak for yourself.

14. Feb 2, 2008

### Gib Z

We want to help, not give you the bloody answer. We're not a freaking oncall service at here for your petty mathematical needs. I speak for the vast majoirty of the forum users here. Only a few people, like you, find it nessescary to ruin and disobey forum rules.

15. Feb 3, 2008

### HallsofIvy

So far you have:

1. Shown that you cannot do even basic arithmetic.