MT2002 Analysis

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## Some properties of convergent sequences

1. Any convergent sequence is bounded (both above and below).

Proof
Suppose that the sequence ( an) . Take = 1 (say). Then choose N so that whenever n > N we have an within 1 of .

Then (apart from the finite set { a1, a2, ... , aN} all the terms of the sequence are bounded by + 1 and - 1.
So an upper bound for the sequence is max {a1 , a2 , ... , aN , + 1 }. Similarly one can find a lower bound.

2. Arithmetic properties

If the sequence (an) and (bn) then

(i) (an+ bn) +
(ii) (an- bn) -
(iii) (anbn)
(iv) (an/bn) / (provided bn 0 and 0).

Proof
Given > 0, choose n1such that if n > n1 then an is closer than to and choose n2 so that when n > n2 we have bn is closer than to .

Then if n > max {n1, n2} we have an+ bn is closer than 2 to + and an- bn is closer than 2 to - and so is arbitrarily small.
The others are a bit trickier:

|anbn-| | (an- )bn+ (bn- )| |bn| |(an- | + || |bn- | |bn| + || .
But (bn) is bounded by Property 1 above and so this is < M for some bound M and so can be made arbitrarily small by choosing small enough.

To prove the result about quotients, we first prove that (1/bn) 1/ and then use (iii) above.
|1/bi - 1/| = | - bi| < /|bi|.
So if we could prove that 1/|bi| is bounded then we would be OK.
Now 1/|| is OK so we've only got to worry about (1/bn).
We are told that (bn) and so for n large enough ( > N say) we have bn is within |/2 | of and hence is at least |/2| away from 0.
But then 1/bn < 1/|/2| and so is bounded as required.

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JOC September 2002