Monotone Convergence Theorem (Real Analysis)/Examples/n - 1 over n

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Example of Use of Monotone Convergence Theorem (Real Analysis)

The sequence $\sequence {a_n}_{n \mathop \ge 1}$ defined as:

$a_n = \dfrac {n - 1} n$

is convergent to the limit $1$.


Proof

From Set of Numbers of form n - 1 over n is Bounded Above, $\sequence {a_n}$ is bounded above with supremum $1$.

Then we have that:

\(\ds \forall n \in \N_{\ge 2}: \, \) \(\ds a_{n + 1} - a_n\) \(=\) \(\ds \dfrac {\paren {n + 1} - 1} {n + 1} - \dfrac {n - 1} n\)
\(\ds \) \(=\) \(\ds \dfrac n {n + 1} - \dfrac {n - 1} n\)
\(\ds \) \(=\) \(\ds \dfrac {n^2 - \paren {n + 1} \paren {n - 1} } {\paren {n - 1} n}\)
\(\ds \) \(=\) \(\ds \dfrac {n^2 - \paren {n^2 - 1} } {\paren {n - 1} n}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \dfrac 1 {\paren {n - 1} n}\) simplifying
\(\ds \) \(>\) \(\ds 0\) as both $n > 0$ and $n - 1 > 0$


When $n = 1$ we have:

\(\ds a_{n + 1} - a_n\) \(=\) \(\ds \dfrac {\paren {1 + 1} - 1} {1 + 1} - \dfrac {1 - 1} 1\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 - \dfrac 0 1\)
\(\ds \) \(=\) \(\ds \dfrac 1 2\)
\(\ds \) \(>\) \(\ds 0\)

So:

$\forall n \in \N_{>0}: a_{n + 1} - a_n > 0$

Thus $\sequence {a_n}$ is strictly increasing.

The result follows from the Monotone Convergence Theorem (Real Analysis).

$\blacksquare$


Sources

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