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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).`
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