Series- convergent or divergent?

In summary, the conversation discusses the convergence or divergence of the series from n=1 to infinity, n!/n^n. The comparison test and limit comparison test are mentioned, but it is noted that the series does not qualify as a geometric series. The use of logarithms and the limit test are suggested as potential methods for solving the problem, but the poster requesting help is reluctant to show their own work and instead asks for someone to write out the solution. The conversation also includes insults and a discussion on forum etiquette.
  • #1
nuclearrape66
31
0
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 don't know if the geometric series thrm applies here.

help please.
 
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  • #2
A geometric series has a constant factor between successive elements, so 1/n^n does not qualify. Look at:

[tex] \frac{n!/n^n}{(n-1)!/(n-1)^{n-1}} [/tex]
 
  • #3
statusx...help me solve my problem rather than leading me to different problems

advie to a homework helper...

if geometric series doesn't apply here, what shouudl be done when comparing
 
  • #4
Re-read StatusX's post. The last part of his message is a major hint.
 
  • #5
hmmm what does (n-1)! = ?

n!(n-1)?
 
  • #6
It means take n, subtract 1, then apply the factorial operator to the result.
 
  • #7
nuclearrape66 said:
hmmm what does (n-1)! = ?

n!(n-1)?
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
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 [tex]y[/tex],

[tex]
y &= \lim_{n \rightarrow \infty} \left | \left ( \frac{n}{n+1} \right )^n \right |\\
[/tex]

take the ln of both sides,

[tex]
\ln y &= \lim_{n \rightarrow \infty} n {\ln \left |\frac{n}{n+1}\right | }
[/tex]

then use L'Hopitals Rule, and solve for [tex]y[/tex].
 
  • #9
can anyone write the solution?

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

thanks
 
  • #10
nuclearrape66 said:
can anyone write the solution?

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

thanks

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
oh okay...ratio test. now i see
 
  • #12
nuclearrape66 said:
statusx...help me solve my problem rather than leading me to different problems

advie to a homework helper...

if geometric series doesn't apply here, what shouudl be done when comparing

nuclearrape66 said:
can anyone write the solution?

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

thanks

It seems it wasn't enough to have your own "give me the answer quickly" attitude, but you want to spread it, like in this thread: https://www.physicsforums.com/showthread.php?t=212583

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 there's 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
are you going to go through my inbox?

and what you're doing right there is whoring my thread...if you don't wanan help don't help. dotn speak for everyone, speak for yourself.
 
  • #14
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
So far you have:

1. Shown that you cannot do even basic arithmetic.

2. Insulted people who tried to help you.

3. Refused to even try yourself.

I think we would all be better off if you just went somewhere else.
 
  • #16
stop whoring my thread, if you don't like it, then don't bother to click on my thread and waste your time...
 

Related to Series- convergent or divergent?

What is a series?

A series is a mathematical concept that refers to a sequence of numbers that are added together.

What does it mean for a series to be convergent?

A convergent series is one in which the terms of the series approach a fixed limit as the number of terms increases. In other words, the sum of the terms of the series approaches a finite value as more terms are added.

What does it mean for a series to be divergent?

A divergent series is one in which the terms of the series do not approach a fixed limit as the number of terms increases. This means that the sum of the terms of the series does not approach a finite value and may continue to increase or decrease without bound.

How can I tell if a series is convergent or divergent?

There are several tests that can be used to determine if a series is convergent or divergent, including the ratio test, the root test, and the integral test. These tests involve analyzing the behavior of the terms in the series and comparing them to known patterns of convergence and divergence.

Why is it important to know if a series is convergent or divergent?

Understanding whether a series is convergent or divergent is crucial in many areas of mathematics and science, including calculus, physics, and statistics. It allows us to make accurate calculations and predictions, and to determine the behavior of systems and processes. Additionally, it helps us to understand the fundamental properties of numbers and how they interact with each other.

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