# The sequence xn have 0<x1<1 and x(n+1)=.25((xn)^5+3xn)? is limit x(n+2)/xn=9/16?

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### 5 Answers

Now prove that `lim_(n->oo) x_n=0`

Indeed if it doesn't:

`lim_(n->oo)x_(n+1)/x_n=lim_(n->oo) 1/4(x_n^4+3)`

`x_n` and `x_(n+1)` converge same limit , thus limit on left side is 1.

Since :

`lim_(n->oo) 1/4x_n^4+3/4=1` `lim_(n->oo) 1/4 x_n^4=1/4`

that is `lim_(n->oo) x_n^4=1` absurd. (the absurd is that limit not zero.)

Then:

`lim_(n->oo) x_(n+2)/x_n= 1/16 lim_(x->oo)(x_n^4+3)^2` `=9/16`

Now `x_(n+2)= 1/4 x_(n+1)(x_(n+1)^4+3)`

and: `x_(n+1)= 1/4 x_n(x_n^4+3) `

Since:

`x_(n+2)/x_(n+1)= 1/4 (x_(n+1)^4+3)`

and:

`x_(n+1)/x_n= 1/4 (x_n^4+3) `

and :

`x_(n+2)/x_n= x_(n+2)/x_(n+1) xx x_(n+1)/x_n` `=(x_(n+1)^4+3)/(x_n^4+3)`

Let's show now that `x_n` decreases:

Indeed: `x_(n+1)-x_n= 1/4 x_n^5- 3/4 x_n - x_n=1/4 x_n(x_n^4-1)`

Since `x_n <1` `x_(n+1)-x_n<0`

`x_n=1/4 x_n(x_n^4+3)`

`0<x_n <1` indeed, `0<x_1<1` `x_2=1/4 x_1(x_1^4+3)`

Since `x_1 <1` `(3+x_1) <4` so that :

`x_2 <1/4 x_1 xx 4 <1/4 xx4 =1`

further, being true for `x_n` :

`x_(n+1)=1/4 x_n(x_n^4 +3)<1`

We have given sequence

`{x_n}` ,where `0<x_1<1 , and x_(n+1)=.25((x_n)^5+3x_n)`

`Thus`

`x_(n+2)=.25((x_(n+1))^5+3x_(n+1))`

`(x_(n+2)/x_n)=.25((x_(n+1))^5+3x_(n+1))/x_n`

``

Thus we can write

`lim_(n->oo){x_n}=lim_(n->oo){x_(n+1)}`

Let `lim_(n->oo)x_n=l`

therefore

`lim_{n->oo}x_(n+1)=lim_{n->oo}.25((x_n)^5+3x_n)`

`l=.25(l^5+3l)`

`1/.25-3=l^4`

`l^4=1`

`therefore0<x_1<1`

`l=1`

`lim_{n->oo}{x_(n+2)/x_n}=1!=9/16`