Combination Theorem for Sequences/Real/Sum Rule

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Theorem

Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.

Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:

$\displaystyle \lim_{n \mathop \to \infty} x_n = l$
$\displaystyle \lim_{n \mathop \to \infty} y_n = m$


Then:

$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$


Proof

Let $\epsilon > 0$ be given.

Then $\dfrac \epsilon 2 > 0$.

We are given that:

$\displaystyle \lim_{n \mathop \to \infty} x_n = l$
$\displaystyle \lim_{n \mathop \to \infty} y_n = m$

By definition of the limit of a real sequence, we can find $N_1$ such that:

$\forall n > N_1: \size {x_n - l} < \dfrac \epsilon 2$

where $\size {x_n - l}$ denotes the absolute value of $x_n - l$

Similarly we can find $N_2$ such that:

$\forall n > N_2: \size {y_n - m} < \dfrac \epsilon 2$


Let $N = \max \set {N_1, N_2}$.

Then if $n > N$, both the above inequalities will be true:

$n > N_1$
$n > N_2$


Thus $\forall n > N$:

\(\displaystyle \size {\paren {x_n + y_n} - \paren {l + m} }\) \(=\) \(\displaystyle \size {\paren {x_n - l} + \paren {y_n - m} }\)
\(\displaystyle \) \(\le\) \(\displaystyle \size {x_n - l} + \size {y_n - m}\) Triangle Inequality for Real Numbers
\(\displaystyle \) \(<\) \(\displaystyle \frac \epsilon 2 + \frac \epsilon 2\)
\(\displaystyle \) \(=\) \(\displaystyle \epsilon\)


Hence the result:

$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$

$\blacksquare$


Also see


Sources