# Series `x_(n+1)=(x_(n)^5+3x_(n))/4` show this `0<x_(n)<1`

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### 1 Answer

Notice that the series is positive for all `x > 0` and you need to prove that any term of the series is less than 1.

Considering `x_n = 1 ` and substituting 1 for `x_n` in the given reccurence relation yields:

`x_(n+1) = (1^5 + 3*1)/4 => x_(n+1) = (1+3)/4`

`x_(n+1) = 4/4 => x_(n+1) = 1`

**Since `x_(n+1)` is larger that `x_n` , then `x_n < ` 1, hence `x_n` `in (0,1).` **