Question

Got a very hard one (off my head, but probably exists).

Answer 1

\[\sum_{n=1}^{\infty}\frac{n!}{n^n}\]

Answer 2

ratio test will do

Answer 3

ratiooo

Answer 4

if you don't like ratio test, here is alternative by comparison test :

\[\frac{n!}{n^n}=\frac{1}{n}\frac{2}{n}\cdots\frac{n-1}{n}\frac{n}{n}<\frac2{n^2}\]

Answer 5

What is n! (factorial of n) doing in this situation ?

n! = n * (n - 1) * (n-2) * ... 1
(for n number of terms)

n^n = n * n * n * n
(for n number of terms)

OR, same way
n! = n * (n-1) * ... * 2 (the 1 doesn't matter)

and n! =< n^(n-1) (for any n)

So we take the limit

lim n^(n-1) / n^n = 0
n -> INFTY

Answer 6

So series converges

Answer 7

\[\sum_{n=1}^{\infty}\frac{n!}{n^n}\] is compared to\[\sum_{n=1}^{\infty}\frac{n^{n-1}}{n^n}\]

Answer 8

because\[n! \le n^{n-1}\] for any n

Answer 9

that shows the limit of sequence converges

this is not sufficient for "series" to converge

Answer 10

you will have to use ratio test or comparison test

Answer 11

\[\sum_{n=1}^{\infty} \frac{ 2 }{ n^2 } \implies converges (p=2>1)\] just use comparison test the way rational showed you

Answer 12

I just didn't get why that is\[<\frac{2}{n^2}\]

Answer 13

\[\frac{n!}{n^n}=\frac{1}{n}\frac{2}{n}\cdots\color{blue}{\frac{n-1}{n}}\frac{n}{n} \]

do you agree that blue term is less than 1 ?

Answer 14

yes, I do

Answer 15

what about each term between 2/n and n/n

Answer 16

all of them are less than 1, agree ?

Answer 17

\[\dfrac{a}{n}\]

is less than 1 when \(a\lt n\)

Answer 18

well, besides of n/n, but all of them are equal to or less than 1

Answer 19

each of the terms between 2/n and n/n is "strictly" less than 1.

(not including n/n)

Answer 20

yes

Answer 21

so they product will be also "strictly" less than 1, yes ?

Answer 22

yes, A * B < 1, if {A,B}<1

Answer 23

\[\begin{align}\frac{n!}{n^n}&=\frac{1}{n}\frac{2}{n}\cdots\color{blue}{\frac{n-1}{n}}\frac{n}{n}\\~\\&\lt \frac{1}{n}\frac{2}{n}\cdot 1\\~\\&=\dfrac{2}{n^2}\end{align}\]

Answer 24

I am here....) glitching

Answer 25

yes

Answer 26

I was about to put up a reply and something happened. here is what I was going to say.

Answer 27

yes, for the first 2 and the last term we get 2/n^2

and when you multiply this by more terms (which any of them is less than 1) you get 2/n^2 decreased.

so 2/n^2 times (less than 1) you get less than 2/n^2

Answer 28

poor wording sorry.

Answer 29

and we are using that as out comparison I get it.
and this way we can tell it converges....

Answer 30

but is there a way to find the value it converges to (without just putting the series in wolfram) ?

Answer 31

not so easy this time as it wont telescope

Answer 32

yes, but\[2\sum_{2}^{\infty}\frac{1}{n^2}\]\[1/2+1/4=3/4\]\[3/4+1/8=7/8\]\[7/8+1/16=15/16\]

Answer 33

I can tell it converges by simply writing some partial sums

Answer 34

well, it will always end up summing to n/(n+1)

Answer 35

we can test partial sums always. if you're allowed to use p-series test, you can simply say :

\(\sum\limits_{n=1}^{\infty}\dfrac{1}{n^2}\) converges by p-series test

Answer 36

yes, it is bound.

Answer 37

but, is there a way to know what \(\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^n}\) converges too ?

Answer 38

to*

Answer 39

i would evaluate first few partial sums then i get bored and move on to next problem

Answer 40

in exams you will never be asked to compute the value it converges to.. unless there is an obvious way to find it

Answer 41

yes, just wanted to know how... anyway, I guess I am going off-line right now:)

Answer 42

tnx for your time and effort!

Answer 43

yw here is a nice way to work it using ratio test
http://math.stackexchange.com/questions/1196708/convergence-of-sum-n-fracnnn