Question

Let \(y_1:=\sqrt p\), p > 0, \(y_{n+1}=\sqrt {p +y_n}\), \(\forall n \in N\)
Show \(\{y_n\}\) converges and find limit.
Please, help

Answer 1

To show \(\{y_n\}\) converges, we need show it monotone and bounded.

Answer 2

\(y_2 = \sqrt{p+y_1}\\y_2-y_1=\sqrt{p+y_1}-y_1\)

Answer 3

@phi @ganeshie8

Answer 4

@hartnn @myininaya

Answer 5

\(p>p+\sqrt p\\\sqrt p>\sqrt{p+\sqrt p}\)
\(\sqrt{p+\sqrt p}- \sqrt p<0\)
replace \(\sqrt p =y_1\)
we have \(\sqrt {p+ y_1}-y_1<0\)
and \(y_2 =\sqrt{p+y_1}\)
so that \(y_2 -y_1 <0\\y_2<y_1\)
\(\{y_n\}\) decreases for n=1,2

Answer 6

oh, I messed up, < ,not > so that , it is increasing for n = 1,2 . I am sorry.

Answer 7

yeah i was gonna say that lol

Answer 8

But whatever, What I need is induction step for n = k+1

Answer 9

whats your upperbound ?

Answer 10

you need to show that the sequence is bounded first, right ?

Answer 11

I don't know. Now I just show it converges, after then find bound by other process. That process is .... a mess.

Answer 12

No, @ganeshie8 I have to prove it converges first, from then, I can assume that it is bounded and find out the bound later.

Answer 13

And that is lim,

Answer 14

{y_n} converges if it is :
1) bounded
2) monotone

Answer 15

Ok, if so, what should I have to do? find bounds, right?
btw, I got induction step. :) hihihi
Now I find bound

Answer 16

for bounded, show that \(\large y_n \lt L\) for all \(\large n\)
for monotone, show that \(\large y_n \lt y_{n+1}\) for all \(\large n\)

Answer 17

let me go through your induction step again, you're trying to prove the sequence is increasing right ?

Answer 18

yes,

Answer 19

just repeat the step again with \(y_{k+2} \) and \(y_{k+1}\). hehehe.

Answer 20

Now, find bounded:
Let x = lim \(y_n\) then x = lim \(y_{n=1}\) also ( while x --> infty\)
so that \(x = \sqrt{p +x}\)
square both sides
\(x^2 = p+x\)
\(x^2-x-p=0\)
\(x= \dfrac{1\pm \sqrt{1+4p}}{2}\)
since 1+4p> 1 for all p>0, we reject the minus value which leads x to negative number
so that the bounds of \(\{y_n\}\) is \(\dfrac{1+\sqrt{1+4p}}{2}\)

Answer 21

for example you want to show that sqrt( 4 + sqrt( 4 + sqrt( 4 + ... )))) converges

Answer 22

I don't get what you mean

Answer 23

your question is asking the limit of square roots, infinite square rooting

Answer 24

if we let p = 4 , for example

Answer 25

Nope, my question is a proving one.

Answer 26

After proving, find the limit.

Answer 27

both parts look good to me !

Answer 28

yyyyyyyyes! Thanks for comfirming.

Answer 29

i was catching up *

Answer 30

\[\lim\limits_{n\to \infty} y_{n+1} = \lim\limits_{n\to \infty} y_{n} = L \implies L = \sqrt{p+L} \implies L = \dfrac{1+\sqrt{1+4p}}{2}\]

Answer 31

also you can mention that it is bounded below by zero, since all the terms are positive and adding /square rooting is positive

Answer 32

you have proved that the sequence is increasing in your first step itself which is bit unusual as you could have used the upper bound in proving the sequence is monotone also. but your logic sounds great !

Answer 33

Show me your stuff, please @ganeshie8 . I need know as many method as possible.

Answer 34

Notice that the work for upper bound is independent of monotone part

Answer 35

im going over your proof, {yn} is actually an increasing sequence

Answer 36

Having known that the sequence is bounded above by \(\large L\), we go ahead and prove that the sequence is increasing using induction

Answer 37

where did you show the sequence is bounded above by L ?

Answer 38

in the earlier limit

Answer 39

I must have missed it, i dont see an upper bound