Question

Prove the following Limit exist and evaluate.
Lim n->∞ [ (1+1/n)(1+2/n)...(1+n/n)]^(1/n)

Answer 1

Hint : try to evaluate first \[\ln(U_n) \]

Answer 2

\[U_{n}=\left[ (1+1/n)(1+2/n)...(1+n/n)\right]^{1/n}\]

Answer 3

ok so I know that lim n->∞ (1+a/n)^n = e^a can I use this? does it still hold for when the exponent is (1/n)?

Answer 4

No ! It''s not that ! Riemann Sum ? does it mean smthing to you ?

Answer 5

Yes its how we approximate the area under the curve and compute the integral

Answer 6

okk ! you know this
\[\lim_{n \rightarrow \infty} \frac{b-a}{n}\sum_{k=1}^{n}f(a+k \frac{b-a}{n})=\int\limits_{a}^{b}f(t)dt\]

Answer 7

yes

Answer 8

lol its cool

Answer 9

now Compute ln(U_n) see what you get !

Answer 10

I took the ln(Un) and got 2/n? 1/n ln [ (1+1\n)(1+2/n)...(1+n/n)] => ln [(1/n+1/n²) (1/n+2/n²) ... (1/n+n/n²) ]

Answer 11

then (1/n +n/n²) = 2/n ? then as n gets bigger it tends toward 0?

Answer 12

wait !

Answer 13

ln=Log

Answer 14

use this
ln(a.b.c.d.e.)=ln(a)+ln(b)+ln(c)+ln(d)+ln(e)

Answer 15

ah crap why did I forget that :/

Answer 16

np at all ! It happens all the time ! :)

Answer 17

so then.
[ln(1+1/n) + ln (1 + 2/n) + ln (1 + n/n) ]/n

Answer 18

yes ! true ! Try to use Riemann...

Answer 19

choose a,b and f !

Answer 20

are you there ?

Answer 21

yes I was thinking a and b are the starting and end points so a=1 and b=∞

Answer 22

I think then what is f...

Answer 23

is it the um 1/n ln (1 + n/n)? because it is a function?

Answer 24

then I would just integrate?

Answer 25

why b infiniti ! Think again ! and re-write what you got ln(U_n)...with Sigma and "k" "n"

Answer 26

like I did Exactly !

Answer 27

don't choose a,b ot f...just re-write it using \[\sum_{k=1}^{n}\]

Answer 28

ln ∑k=1n ln(1+k/n)

Answer 29

how did you get the sigma to show up like that?

Answer 30

loool ! use your equation editor

Answer 31

1/n sigma k=1 ln(1+k/n)

Answer 32

yes ! Good ! Try to choose f a and b to be just like what I typed !

Answer 33

\[1/n \sum_{k}^{\infty} \ln (1 + k/n)\]

Answer 34

not infinity ! from where did you get infinity ! :) :)

Answer 35

well I figured the interval starts at 1 and goes from 1 to 2 to n which approches infinity

Answer 36

1 to 2; then 2 to 3; ... all the way to n

Answer 37

maybe b=n?

Answer 38

no

Answer 39

what ! a and b are independant to n ! Just.. think ! re-write it correctly ! "n" is not changing now ! It's like I've choosed it ! but you don't know its exact value ! all I'm asking to re-type it correctly :) ! use the equation editor !

Answer 40

\[1/n \sum_{k=1}^{n} \ln (1+k/n)\]

Answer 41

oh lol I see now it was kinda small and I thought you had infinity at the top of the sigma for some reason

Answer 42

:) :) a big Attention is sometimes helpful !
now compare it ! to the expression ! then choose a,b and f !

Answer 43

what is your choice ?!

Answer 44

a=1 b=2 f= ln(1+k/n)

Answer 45

f(t) =...... f is independ to n and k but...what you're seeing "inside" sigma is just the image ! of something depending to n,k

Answer 46

hum..

Answer 47

tell me !

Answer 48

im not sure what f is

Answer 49

i mean f(t)

Answer 50

I'll give you an example okk !

Answer 51

ok

Answer 52

choose f a and b
\[f(a+k \frac{b-a}{n} )=(1+\frac{2k}{n})^2\]

Answer 53

a=1 b=3

Answer 54

f= (1+ 2k/n)?

Answer 55

I cant evaluate the function if I dont know k and n

Answer 56

\[f(1+k \frac{3-1}{n})=(1+k \frac{3-1}{n})^2\]
like I'm asking you \[f(x_n)=x_n^2\]
what f can be ?

Answer 57

x? its a squaring fuction

Answer 58

you mean \[f(t)=t^2 ? yeaah ?\]

Answer 59

f(x) =x^2

Answer 60

yes sorry thats what i ment my internet dropped for a sec

Answer 61

np ... let's go back to your question ! if that notation x_n worked for you ! re-do it ! with your question !

Answer 62

oh yeah so is my f = ln(x)?

Answer 63

yes ! true ! what is the lim ?!

Answer 64

+ infinity? because the log just grows bigger?

Answer 65

this is the full expression \[\lim_{b \rightarrow \infty} 1/n \sum_{k=1}^{n} \ln ( 1 + k/n)\]

Answer 66

so then the limit is 0?

Answer 67

see the expression ! that limite is equal to = ... (you choosed a and b ...what for )

Answer 68

a=1 b=2 or are you asking why i put numbers in for a and b?

Answer 69

no ! I'm just want to be sure if you know what do you want exactly ! what you're really doing !

Answer 70

im trying to find the limit to the problem? maybe im getting lost

Answer 71

Could i just take the integral over the interval a and b and get the answer?

Answer 72

so what ?! \[\lim_{n \rightarrow \infty} \ln(U_n)=\lim_{n \rightarrow \infty} \frac{2-1}{n} \sum_{k=1}^{n}f(1+k \frac{2-1}{n})=\int\limits_{1}^{2}f(x)dx\]
you had a b and f !

Answer 73

where fx would = ln(x) ir \[\int\limits_{1}^{2} \ln(x)dx ?\]

Answer 74

Yess...you now how to evaluate that !

Answer 75

1/2?

Answer 76

No ! Integration by parts !

Answer 77

sorry it has been a second since i have integrated give me a moment

Answer 78

:D take your time just gimme a correct answer !

Answer 79

log(4)-1

Answer 80

correct ! that was \[\lim_{n \rightarrow \infty}\ln(U_n)=\ln(4)-1\]
now ! \[\lim_{n \rightarrow \infty}U_n\]=??

Answer 81

\[U_n=\exp(\ln(U_n))\]??

Answer 82

sorry my internet went down
4/e

Answer 83

Wow that took forever but how did you know to first take the log of it?

Answer 84

Sorry I got that answer like a few seconds after you posted the question 16 mins ago my internet just went down :( but Thank You!!!!

Answer 85

Np ! and for your question ! I don't know...( intuition+experience.....)Maybe...!
Just do more exercises...and you'll be fine !

Answer 86

ah ok thanks I appreciate it :)

Answer 87

and I believe that \[\sum_{}^{}\] is more easier than \[\prod_{}^{}\]
you're welcome :) ! wish me luck I really need it !

Answer 88

it is lol good luck!