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

oldnick | Student

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.)


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

oldnick | Student

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) `  


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


`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)`



oldnick | Student

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`

oldnick | Student

`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`  

pramodpandey | Student

We have given sequence

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






Thus we can write  


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









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