Some properties of convergent sequences

1. 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 within 1 of α.

   Then (apart from the finite set { a₁, a₂, ... , 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 {a₁ , a₂ , ... , a_N , α + 1 }. Similarly one can find a lower bound.
   
   
   

2. 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_nb_n)→ α β
   

   (iv) (^a_n/_bn)→ ^α/_β (provided b_n≠ 0 and β ≠ 0).
   

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

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

   |a_nb_n-αβ| ≤ | (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 < 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 - 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.