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

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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.

$\blacksquare$


Sources