# Convergent Real Sequence/Examples/x + x^n over 1 + x^n

## Example of Convergent Real Sequence

The sequence $\sequence {a_n}$ defined as:

$a_n = \dfrac {x + x^n} {1 + x^n}$

is convergent for $x \ne -1$.

Then:

$\displaystyle \lim_{n \mathop \to \infty} \dfrac {x + x^n} {1 + x^n} = \begin {cases} 1 & : x \ge 1 \\ x & : -1 < x < 1 \\ 1 & : x < -1 \\ \text {undefined} & : x = -1 \end {cases}$

## Proof

Let $\size x < 1$.

We have:

 $\displaystyle \dfrac {x + x^n} {1 + x^n}$ $\to$ $\displaystyle \dfrac {x + 0} {1 + 0}$ $\displaystyle \text {as n \to \infty}$ Sequence of Powers of Number less than One $\displaystyle$ $=$ $\displaystyle x$

Now let $\size x > 1$.

We have:

 $\displaystyle \dfrac {x + x^n} {1 + x^n}$ $=$ $\displaystyle \dfrac {\dfrac 1 {x^{n - 1} } + 1} {\dfrac 1 {x^n} + 1}$ dividing top and bottom by $x_n$ $\displaystyle$ $=$ $\displaystyle \dfrac {0 + 1} {0 + 1}$ $\displaystyle \text {as n \to \infty}$ Sequence of Powers of Reciprocals is Null Sequence $\displaystyle$ $=$ $\displaystyle 1$

Then when $x = 1$:

 $\displaystyle \dfrac {x + x^n} {1 + x^n}$ $=$ $\displaystyle \dfrac {1 + 1^n} {1 + 1^n}$ $\displaystyle$ $=$ $\displaystyle \dfrac {1 + 1} {1 + 1}$ $\displaystyle$ $=$ $\displaystyle 1$

When $x = -1$:

 $\displaystyle \dfrac {x + x^n} {1 + x^n}$ $=$ $\displaystyle \dfrac {\paren {-1} + \paren {-1}^n} {1 + \paren {-1}^n}$ $\displaystyle$ $=$ $\displaystyle \dfrac {-1 + 1} {1 + 1}$ for even $n$ $\displaystyle$  $\displaystyle \dfrac {-1 + -1} {1 + -1}$ for odd $n$ $\displaystyle$ $=$ $\displaystyle 0$ for even $n$ $\displaystyle$  $\displaystyle \text {undefined}$ for odd $n$

Hence the result.

$\blacksquare$