Some properties of convergent sequences

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

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

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

2. Arithmetic properties

   If the sequence (a_n) and (b_n) then
   

   
   

   (i) (a_n+ b_n) +
   

   (ii) (a_n- b_n) -
   

   (iii) (a_nb_n)
   

   (iv) (^a_n/_bn) / (provided b_n 0 and 0).
   

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

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

   |a_nb_n-| | (a_n- )b_n+ (b_n- )| |b_n| |(a_n- | + || |b_n- | |b_n| + || .
   But (b_n) 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/| = | - b_i| < /_|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 (b_n) and so for n large enough ( > N say) we have b_n 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.