Sums of sequences

1. Properties of the notion of limit with addition

So far in this course you have seen several sequences that converge with proofs of their convergence. These include:

  • The sequence a n = 1 n which converges to 0 as n .
  • The sequence a n = x n (for a fixed number 0 < x < 1 ) which also converges to 0 as n .
  • The constant sequence a n = c which converges to c as n .

We want to build up our repertoire of such sequences, and we start by looking at addition.

Theorem.

Let ( a n ) and ( b n ) be convergent sequences with limits l and m respectively. Let the sequence ( c n ) be defined by c n = a n + b n . Then c n l + m as n .

Proof.

We are given that ε > 0   N   n   n N | a n - l | < ε and ε > 0   N   n   n N | b n - m | < ε and must prove that a n + b n l + m .

Subproof.

Let ε > 0 be arbitrary.

Subproof.

Let N 1 such that n   n N 1 | a n - l | < ε / 2 and N 2 such that n   n N 2 | b n - m | < ε / 2 .

Let N = max ( N 1 , N 2 ) .

Subproof.

Let n be arbitrary.

Subproof.

Assume n N .

Then n N 1 and n N 2 so | a n - l | < ε / 2 and | b n - m | < ε / 2 .

So | ( a n + b n ) - ( l + m ) | | a n - l | + | b n - m | < ε / 2 + ε / 2 = ε by the triangle inequality.

It follows that n N | a n + b n - ( l + m ) | < ε .

So n   n N | a n + b n - ( l + m ) | < ε .

So N   n   n N | a n + b n - ( l + m ) | < ε .

So ε > 0   N   n   n N | a n + b n - ( l + m ) | < ε , as required.

This result enables us to write down the limit of sequences such as 1 + 1 n and x n + y n . It also enables us to compute the limit of 2 a n writing this as a n + a n , and 3 a n , 4 a n , etc. The following result (which will be improved in a later web page) roll all these together in a single proposition.

Proposition.

Let ( a n ) be a convergent sequence with limit l , and let k be a number. Let the sequence ( b n ) be defined by b n = k a n . Then b n k l as n .

Proof.

We are given that ε > 0   N   n   n N | a n - l | < ε and b n = k a n for all n . We will assume k 0 , for if k = 0 then b n is the constant sequence 0 which converges to 0 = k l .

Subproof.

Let ε > 0 be arbitrary.

Subproof.

Let N such that n   n N | a n - l | < ε / | k | since k 0 hence ε / | k | > 0 .

Then for each n N , | b n - k l | = | k | | a n - l | < ε , i.e., n   n N | b n - k l | < ε .

So N   n   n N | b n - k l | < ε .

So ε > 0   N   n   n N | b n - k l | < ε , as required.

These two results show a similar theorem also holds for subtraction.

Theorem.

Let ( a n ) and ( b n ) be convergent sequences with limits l and m respectively. Let the sequence ( c n ) be defined by c n = a n - b n . Then c n l - m as n .

Proof.

c n = a n - b n = a n + ( -1 ) b n , so ( -1 ) b n - m by the proposition on multiplying -1 by the sequence b n , and c n = a n + ( -1 ) b n l + ( - m ) = l - m , by the theorem on addition.