Question #159885

Prove that a sequence is bounded if and only if it is both bounded above and bounded below


1
Expert's answer
2021-02-04T06:32:12-0500

Let <<an>> be a sequence. given that <<an>> is bounded, then by definition of boundedness

\exist k \in R such that |an| \leq k \forall n

    \implies -k \leq |an| \leq k \forall n

    \implies -k \leq |an| & |an| \leq k \forall n

which implies that <<an>> is bounded below by -k and bounded above by k.

therefore, <<an>> is both bounded above and bounded below.

Conversely:

now suppose that <<an>> is both bounded above and bounded below.

let L, k \isin R and <<an>> is bounded below by L and bounded above by k for every n

    \implies an \leq k & an\geq L \forall n

\because -|L| - |k| \leq L & k\leq |L| + |k|

then,

if L \leq an \leq k \forall n

    \implies -|L| - |k| \leq L \leq |an| \leq k \leq |L| + |k| \forall n

    \implies -|L| - |k| \leq |an| \leq |L| + |k| \forall n

hence we conclude that <<an>> is bounded by |L| + |k|.

<<an>> is bounded.




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