# 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 $l ∈ ℝ$ and $m ∈ ℝ$ 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 and and must prove that $a n + b n → l + m$.

Subproof.

Let $ε > 0$ be arbitrary.

Subproof.

Let $N 1 ∈ ℕ$ such that and $N 2 ∈ ℕ$ such that .

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

Subproof.

Let $n ∈ ℕ$ 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 .

So .

So , 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 $l ∈ ℝ$, and let $k ∈ ℝ$ 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 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 $N ∈ ℕ$ such that since $k ≠ 0$ hence $ε / | k | > 0$.

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

So .

So , 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 $l ∈ ℝ$ and $m ∈ ℝ$ 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.