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

Suppose that the sequence (*a*_{n})→*α*. Take*ε*= 1 (say). Then choose*N*so that whenever*n*>*N*we have*a*_{n}within1 of *α*.Then (apart from the finite set {

*a*_{1},*a*_{2}, ... ,*a*_{N}} all the terms of the sequence are bounded by*α*+ 1 and*α*- 1.

So an upper bound for the sequence is max {*a*_{1},*a*_{2}, ... ,*a*_{N},*α*+ 1 }. Similarly one can find a lower bound.

**Arithmetic properties***If the sequence*(*a*_{n})→*α**and*(*b*_{n})→*β**then*

- (i)
( *a*_{n}+*b*_{n})→*α*+*β*

- (ii)
( *a*_{n}-*b*_{n})→*α*-*β*

- (iii)
( *a*_{n}*b*_{n})→*α**β*

- (iv)
( ^{an}/_{bn})→^{α}/_{β}(*provided b*≠ 0_{n}*and**β*≠ 0).

**Proof**

Given*ε*> 0, choose*n*_{1}such that if*n*>*n*_{1}then*a*_{n}is closer than*ε*to*α*and choose*n*_{2}so that when*n*>*n*_{2}we have*b*_{n}is closer than*ε*to*β*.Then if

*n*> max {*n*_{1},*n*_{2}} we have*a*_{n}+*b*_{n}is closer than 2*ε*to*α*+*β*and*a*_{n}-*b*_{n}is closer than 2*ε*to*α*-*β*and so is arbitrarily small.

The others are a bit trickier:|

*a*_{n}*b*_{n}-*α**β*| ≤ | (*a*_{n}-*α*)*b*_{n}+*α*(*b*_{n}-*β*)| ≤ |*b*_{n}| |(*a*_{n}-*α*| + |*α*| |*b*_{n}-*β*| ≤ |*b*_{n}| + |*α*|*ε*.

But (*b*_{n}) is bounded by Property 1 above and so this is <*M**ε*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*-*b*_{i}| <^{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 (*b*_{n})→*β*and so for*n*large enough ( >*N*say) we have*b*_{n}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.

- (i)

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