Question

lim n->∞ (1-1/4)*(1-1/9)*...*(1-1/n^2)

Answer 1

@quickstudent @waterineyes @Elsa213 @radar @TheAsker2002 @Yolo4mecuite

Answer 2

r there any answer choices?

Answer 3

hmmm

Answer 4

hm

Answer 5

@Alexander95

Answer 6

0 -.-

Answer 7

maybeenot

Answer 8

1/2?

Answer 9

NOT.

Answer 10

cx

Answer 11

lol

Answer 12

the answer is definately les than 3/4 ill tell u that

Answer 13

^^^^^^

Answer 14

less*

Answer 15

The*

Answer 16

definitely*

Answer 17

I'll*

Answer 18

you*

Answer 19

@Alexander95 is your keyboard working?

Answer 20

that.*

Answer 21

no answer choice but I found out that those expressions can be simplified

Answer 22

umm well

Answer 23

its prollyyyy 1/2 gut feeling

Answer 24

\[\frac{ 3 }{ 4 }*\frac{ 8 }{ 9 }*\frac{ 15 }{ 16}*...*\frac{ n^2-1 }{ n^2 }\]

Answer 25

yeah

Answer 26

but why 1/2 @dan815 instead of 1

Answer 27

itd not 1/2, its a guess, it cant be one since you keeep multiplying by a smaller fraction

Answer 28

hmm lemme tink i think i know a way to solve this

Answer 29

this cud be solve by eigen values

Answer 30

dan815 you're right this product really tends to 1/2

Answer 31

Base is larger is every term, so every term is less than 1, so overall will be more less result.

Answer 32

I.ve multiplied the first 20 terms of this exp
But still need an algebraic solvation of the problem

Answer 33

\[\frac{ 1\times 3 }{ 2\times 2 }*\frac{ 2\times 4 }{ 3\times 3 }*\frac{ 3\times 5 }{ 4\times 4}*...*\frac{ (n-1)\times(n+1) }{ n\times n }\]

Answer 34

then start canceling

Answer 35

pretty cool

Answer 36

\[= \lim_{n \to \infty}\frac{1}{2} \times \frac{n+1}{n}\]

Answer 37

Wow, Zarkon, Great...!! :)

Answer 38

So, Dan was right.. :)

Answer 39

i think i found a cool soln* with eigen values too xD, if u write recurrence relation, prolly too much work for this algebra problem tho