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)→ α. Take ε = 1 (say). Then choose N so that whenever n > N we have an within 1 of α.

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

2. Arithmetic properties

If the sequence (an)→ α and (bn)→ β then

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

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

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

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

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

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JOC September 2001