Question

please help me find this limit


\[\left\{ \frac{n!}{{{n}^{n}}} \right\}\underset{n=1}{\overset{\infty }{\mathop{{}}}}\,\]

Answer 1

Any ideas?

Answer 2

qwll I have an idea but I am pretty sure that I am the only person in his world that is thinking that way

Answer 3

but yes I solved it and its inf

Answer 4

$$\Huge n!=1*2*3*4*...*n$$
$$\Huge n^n=\underbrace{n*n*n*...*n}$$
$$\Huge\text { _________n factors}$$

Answer 5

\[n!=(n+1)(n+2)(n+3)...\]

Answer 6

thats how I did it at least

Answer 7

is it fractional part function?

Answer 8

it seems you were right @skullpatrol indeed I found this at folmram https://www.dropbox.com/s/zq78o61ftgaweyf/Screenshot%202013-12-15%2015.16.12.jpg although it was unclear

Answer 9

but I think both ways are right :D

Answer 10

See? I told you I did it in a paradoxical way I am sure that I don't even know the real approach of these things

Answer 11

Try to use the definitions as much as possible :)

Answer 12

can you please tell me whats your approach ?

Answer 13

@skullpatrol

Answer 14

http://math.info/image/290/ratio_test.gif

Answer 15

@Christos and \(n!\) is not \((n+1)(n+2)(n+3)\cdots\)

Answer 16

and it's not infinity

Answer 17

dude I am trying to find the limit, no test I am allowed to use here

Answer 18

no its infinity

Answer 19

ok sorry

Answer 20

but it's not infinity

Answer 21

Why man?

Answer 22

why is it infinity

Answer 23

that was my result

Answer 24

whats yours?

Answer 25

my result is 0

Answer 26

hm

Answer 27

can you show me your steps so that we can follow

Answer 28

well let me type it in latex.........

Answer 29

ok :)

Answer 30

\[\lim_{n \rightarrow \infty}(n!/n ^{n})\]

Answer 31

lolz @nitz u don't say

Answer 32

\[\lim_{n \rightarrow \infty}((n)(n-1)(n-2)........3.2.1)/(n.n.n..........n)\]

Answer 33

Well instead of thinking about the limit, let's look at something smaller but similar:

5!=1*2*3*4*5
5^5=5*5*5*5*5

In each case you're multiplying 5 numbers together, but the factorial is going to be multiplying smaller numbers while the exponent version is going to make all of the numbers the highest number that the factorial ever reaches.

Answer 34

lolz why don't we wait till she finishes typing in latex

Answer 35

\[\underset{n\to \infty }{\mathop{\lim }}\,\left\{ \frac{1*2*3...(n-1)*n}{n*n*n*n*n....{{n}^{n}}} \right\}_{n=1}^{\infty }\]

Answer 36

and the limit sign and the bracket are redundant

Answer 37

I am a fast typer, just not used in MAthType

Answer 38

What's this (n=1) and (infinity) thing?

Answer 39

@Kainui that denotes a sequence

Answer 40

Please tell mes wrong so that I fix it.

Answer 41

@Kainui and is misused in this scenario

Answer 42

It makes me think this is really some kind of power series. The difference between an infinite sum of these terms and just the limit of one of these terms is the difference between 0 and infinity.

Answer 43

\[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{1*2*3...(n-1)*n}{n*n*n*n*n....{{n}^{n}}} \right]_{n=1}^{\infty }\]

Answer 44

\[\Large\lim_{n\rightarrow\infty}\frac{n!}{n^n}\\\Large=\lim_{n\rightarrow\infty}\frac{1\times2\times\cdots\times n}{n\times n\times\cdots\times n}\\\Large=\lim_{n\rightarrow\infty}\frac1n\times\lim_{n\rightarrow\infty}\frac2n\times\cdots\times\lim_{n\rightarrow\infty}\frac nn\\\Large=0\times0\times\cdots\]

Answer 45

\[\Large=0\]

Answer 46

http://screencast.com/t/W5N6M0kHu

Answer 47

I think that's where you are mistaking

Answer 48

I think that's where you're mistaking @Christos.

Answer 49

the numerator is n! and the denominator is n^n

Answer 50

yes I am talking about the denominator too.

Answer 51

Look, that entire part is n^n. Which is n times itself n times. Just like 3^3 is 3*3*3. and 4^4=4*4*4*4. It's n times itself n times.

Answer 52

n*n*n*n*n..........n^n

Answer 53

No, wrong.

n*n*n*n*...*n = n^n

Answer 54

n*n*n*n*...*n but that kind of presentation implies that you know whats going on inside the "..."

Answer 55

http://screencast.com/t/NZAmQzMMvN

Answer 56

forget about Kainui, I realised were I was wrong. I get it now.

Answer 57

I see

Answer 58

you meant "forget about it Kainui"

Answer 59

No, the "..." represents that we don't know how many n's there are.

2^2=2*2

3^3=3*3*3

4^4=4*4*4*4

5^5=5*5*5*5*5

n^n=n*n*n*...*n

Since n is a variable, we don't know how many times n is being multiplied by itself, so we can't possibly write out n n's!

Answer 60

Ok good =)

Answer 61

yea yea I meant that I understood :D I missed the "it" haha

Answer 62

lolz those two have a great difference in meaning

Answer 63

ironically yea :D look

Answer 64

\[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{{{n}^{n}}*n(cons\tan t)}{n} \right]_{n=1}^{\infty }\]

Answer 65

what's that constant

Answer 66

wait

Answer 67

can u show me the original ques if it's on web

Answer 68

the original questions? https://www.dropbox.com/s/6kpdq04ht4w19u6/Screenshot%202013-12-15%2015.52.14.jpg

Answer 69

which question is that?

Answer 70

past 1. Question (ii)

Answer 71

\[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{{{n}^{n}}-n}{n} \right]_{n=1}^{\infty }\]

Answer 72

oh so there ain't a constant there lol
(btw which question is that coz i can't find that in your list)

Answer 73

That one: http://screencast.com/t/3VIez6jTgrop

Answer 74

I don't see any similarities between these:
http://screencast.com/t/HIz71lvxv5W
http://screencast.com/t/3VIez6jTgrop

Answer 75

\[\underset{n\to \infty }{\mathop{\lim }}\,\left[ n-1 \right]_{n=1}^{\infty }\]

Answer 76

come on don't type latex using the equation button it'll make you v......e.......r.....y...... slow

Answer 77

I am uing MAthType program

Answer 78

if you can't use \(\LaTeX\), then \(\Huge\mbox{don't use it}\)

Answer 79

I instead use latex, to can use it later

Answer 80

anyway learn to type \(\LaTeX\) as a code here

Answer 81

:P

Answer 82

hmm how?

Answer 83

\[\mbox{\[ \lim_{n\rightarrow\infty}\frac{n!}{n^n} \]}\]generates\[\lim_{n\rightarrow\infty}\frac{n!}{n^n}\]

Answer 84

don't bother with it if you can't use it, coz u can still survive with plain text:

limit (n tends to infinity) n!/n^n

Answer 85

I like latex :DDD

Answer 86

limit (n tends to infinity) n!/n^n
=limit (n tends to infinity) (1*2*3*...*n)/(n*n*n*...*n)
=limit (n tends to infinity) (1/n)*(2/n)*(3/n)*... (it won't end)
=limit (n tends to infinity) 0*0*0*...
=0

Answer 87

So we get

\[\left[ \infty -1 \right]-\left[ 1-1 \right]\]

Answer 88

\[\left[ \infty -1 \right]-\left[ 1-1 \right]=\infty -0\]

Answer 89

can you show me how you derived from here:
http://screencast.com/t/3VIez6jTgrop
to here:
http://screencast.com/t/HIz71lvxv5W

Answer 90

ok sure

Answer 91

you deserve a medal @kc_kennylau nice teaching

Answer 92

\[\left\{ \frac{n!}{{{n}^{n}}} \right\}\underset{n=1}{\overset{\infty }{\mathop{{}}}}\,\]




\[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{1*2*3*4...n(n-1)}{n*n*n...{{n}^{n}}} \right]_{n=1}^{\infty }\]

Answer 93

I am sorry correction

Answer 94

\[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{1*2*3*4...n(n-1)}{n*n*n...n} \right]_{n=1}^{\infty }\]



yes yes yes

Answer 95

stop putting the redundant brackets lol

Answer 96

\[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{1*2*3*4...n}{n*n*n...n} \right]_{n=1}^{\infty }\]




now it looks better :P

Answer 97

use stirling's approximation