Question

s.pnh

Answer 1

@ParthKohli

Answer 2

I thought it was supposed to be bounded below but that was wrong

Answer 3

\[ \large a_0=\sqrt{2} \]
\[ \large a_1=\sqrt{2+a_0}=\sqrt{2+\sqrt{2}} \]
\[ \large a_2=\sqrt{2+a_1}=\sqrt{2+\sqrt{2+\sqrt{2}}} \]

Answer 4

Yeah so it gets larger and larger right?

Answer 5

yes

Answer 6

So can't be bounded above

Answer 7

it is actually bounded below

Answer 8

I put that down and it was wrong :(

Answer 9

no it cannot be bounded above

Answer 10

maybe its not bounded at all?

Answer 11

i think it is bounded above. im not sure with the defintions though

Answer 12

Why would it be bounded above?

Answer 13

\[ \large 2+a_0=2+\sqrt{2} \]
\[ \large a_1=\sqrt{2+a_0}\leq\sqrt{2}+\sqrt{a_0}<\sqrt{a_0}=\sqrt{\sqrt{2}}=\sqrt[4]{2}<\sqrt{2} \]

Answer 14

as k->infiniti, the value of term approaches 2

Answer 15

I'm still not sure what the right answer would be

Answer 16

so
\[ \large a_1\leq a_0 \]

Answer 17

yeah

Answer 18

sorry i gotta go
check if \(a_2\leq a_1\) then u will get more information

Answer 19

is this correct definition for bounded above :
any term in the sequence must be <= upper bound

if so, then the sequence is bounded above. because, no term in the sequence can go above the value 2. again im not sure with these definitions, i never did these... .

Answer 20

Bounds mean it wont pass that number so its kind of like an asymptote

Answer 21

why cant any term go past 2?

Answer 22

okay then it is bounded above :

let the the last term of sequence be y,

\(y = \sqrt{2+\sqrt{2+\sqrt{2+........ }}}\)

Answer 23

Okay

Answer 24

if you solve this, you would get y = 2

Answer 25

and how do you prove that?

Answer 26

that expression can be written as :

\(y = \sqrt{2+y}\)

Answer 27

square both sides, and solve the quadratic

Answer 28

how would i write out the series?

Answer 29

didnt get u

Answer 30

I want to plug this in to wolfram to make sure it doesnt go above 2 so how would i do that?

Answer 31

http://www.wolframalpha.com/input/?i=y+%3D+sqrt%282%2By%29&dataset=

Answer 32

oh you wanto plugin the sequence is it

Answer 33

yeah

Answer 34

idk... let me try once though...

Answer 35

ok

Answer 36

no success wid wolfram. but im confident about the last term of the given sequence. it wont cross 2. trust me !

Answer 37

but how can I prove that?!

Answer 38

i have given the proof above. you didnt understand it is it

Answer 39

i dont think that is proof. i think you have to do a series test.

Answer 40

this is a increasing sequence,

last term of the sequence :
\( y = \sqrt{2+\sqrt{2+\sqrt{2+........ }}} \)

Answer 41

oh im not familiar wid those, i thought if we just prove last term of sequence wont go beyond certain value, that is a fair enuf thing to conclude its bounded above

Answer 42

i dont think so

Answer 43

ok may i knw what series test... if u hav time

Answer 44

well there are many tests to do but I first need to know how to write the series. if you can give me that ill show you how to do it

Answer 45

\[\sum_{n=}^{infiniti} something here \]
something like that