Analysis I MT22, Cauchy sequences

Created: January 08, 2023 | Read markdown | About these notes | View in context | Study these flashcards |

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@Define, in full, what it means for a sequence $(a _ n)$ to be Cauchy.

cauchy-sequence-definition
\[\forall \varepsilon > 0 \text{ } \exists N\in\mathbb{N} \text{ s.t.} \forall m > n > N \text{ } \vert a _ m - a _ n \vert < \varepsilon\]

When proving that if $(a _ n)$ is Cauchy, it is bounded, you know that

\[\forall \varepsilon > 0 \text{ } \exists N\in\mathbb{N} \text{ s.t.} \forall m > n > N \text{ } \vert a _ m - a _ n \vert < \varepsilon\]

what value of epsilon $\varepsilon$ and $m$ do you take to show that all terms $n > N$ are bounded?

cauchy-implies-bounded-epsilon-choice

$\varepsilon = 1$, $m = N+1$, $ \vert a _ n - a _ {N+1} \vert < 1$

When proving the forward direction of Cauchy criterion for convergence,

A sequence converges if and only if it is Cauchy.

what magic triangle inequality transformation do you make to $ \vert a _ n - a _ m \vert $?

cauchy-criterion-forward-triangle
\[\begin{aligned} \vert a _ n - a _ m \vert &= \vert a _ n - a + a - a _ m \vert \\ &\le \vert a _ n - a \vert + \vert a - a _ m \vert \end{aligned}\]

When proving the backwards direction of Cauchy criterion for convergence,

A sequence converges if and only if it is Cauchy.

it would be very useful to have something you actually know the sequence converges to. How do you know that the sequence has a convergent subsequence?

cauchy-criterion-backward-subsequence
  • A sequence being Cauchy implies the sequence is bounded.
  • Any bounded sequence has a convergent subsequence.

When proving the backward direction of Cauchy criterion for convergence,

A sequence converges if and only if it is Cauchy.

What magic triangle inequality transformation do you make to $ \vert a _ n - L \vert $?

cauchy-criterion-backward-triangle
\[\begin{aligned} \vert a _ n - L \vert &= \vert a _ n - a _ {b _ {n+1}\space}+a _ {b _ {n+1}\space}-L \vert \\ &\le \vert a _ n - a _ {b _ {n+1}\space} \vert + \vert a _ {b _ {n+1}\space}-L \vert \end{aligned}\]


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