Combination Theorem for Sequences

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[edit] Theorem

Let X be one of the standard number fields \Q, \R, \C.


Let \left \langle {x_n} \right \rangle and \left \langle {y_n} \right \rangle be sequences in X.

Let \left \langle {x_n} \right \rangle and \left \langle {y_n} \right \rangle be convergent to the following limits:

\lim_{n \to \infty} x_n = l, \lim_{n \to \infty} y_n = m


Let \lambda, \mu \in X.


Then the following results hold:


[edit] Sum Rule

\lim_{n \to \infty} \left({x_n + y_n}\right) = l + m


[edit] Multiple Rule

\lim_{n \to \infty} \left({\lambda x_n}\right) = \lambda l


[edit] Combined Sum Rule

\lim_{n \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m


[edit] Product Rule

\lim_{n \to \infty} \left({x_n y_n}\right) = l m


[edit] Quotient Rule

\lim_{n \to \infty} \frac {x_n} {y_n} = \frac l m, provided that m \ne 0


[edit] Proof

[edit] Proof of Sum Rule

We need to show that \lim_{n \to \infty} \left({x_n + y_n}\right) = l + m.

Let ε > 0 be given. Then \frac \epsilon 2 > 0.

Since \lim_{n \to \infty} x_n = l, we can find N1 such that \forall n > N_1: \left|{x_n - l}\right| < \frac \epsilon 2.

Similarly, since \lim_{n \to \infty} y_n = m, we can find N2 such that \forall n > N_2: \left|{y_n - m}\right| < \frac \epsilon 2.

Now let N = \max \left\{{N_1, N_2}\right\}.

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

Thus \forall n > N:

\left|{\left({x_n + y_n}\right) - \left({l + m}\right)}\right| = \left|{\left({x_n - l}\right) + \left({y_n - m}\right)}\right|                    
\le \left|{x_n - l}\right| + \left|{y_n - m}\right|          Triangle Inequality          
< \frac \epsilon 2 + \frac \epsilon 2 = \epsilon                    

Hence \lim_{n \to \infty} \left({x_n + y_n}\right) = l + m.

\blacksquare


[edit] Proof of Multiple Rule

Let ε > 0.

We need to find N such that \forall n > N: \left|{\lambda x_n - \lambda l}\right| < \epsilon.

If λ = 0 the result is trivial.


So, assume \lambda \ne 0.

Then \left|{\lambda}\right| > 0 from the definition of the modulus of λ.

Hence \frac {\epsilon} {\left|{\lambda}\right|} > 0.


We have that x_n \to l as n \to \infty.

Thus it follows that \exists N: \forall n > N: \left|{x_n - l}\right| < \frac {\epsilon} {\left|{\lambda}\right|}.

That is, \forall n > N: \left|{\lambda}\right| \left|{x_n - l}\right| < \epsilon.


But we have:

\left|{\lambda}\right| \left|{x_n - l}\right| = \left|{\lambda \left({x_n - l}\right)}\right|          Modulus of Product          
= \left|{\lambda x_n - \lambda l}\right|                    

Hence \lim_{n \to \infty} \left({\lambda x_n}\right) = \lambda l.

\blacksquare


[edit] Proof of Combined Sum Rule

We need to show that \lim_{n \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m.

From the Multiple Rule above, we have:

  • \lim_{n \to \infty} \left({\lambda x_n}\right) = \lambda l
  • \lim_{n \to \infty} \left({\mu y_n}\right) = \mu m

The result now follows directly from the Sum Rule above:

\lim_{n \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m

\blacksquare


[edit] Proof of Product Rule

We need to show that \lim_{n \to \infty} \left({x_n y_n}\right) = l m.

Since \left \langle {x_n} \right \rangle converges, it is bounded by Convergent Sequence is Bounded.

Suppose \left|{x_n}\right| \le K for n = 1, 2, 3, \ldots.

Then:

\left|{x_n y_n - l m}\right| = \left|{x_n y_n - x_n m + x_n m - l m}\right|                    
\le \left|{x_n y_n - x_n m}\right| + \left|{x_n m - l m}\right|          Triangle Inequality          
\le \left|{x_n}\right| \cdot \left|{y_n - m}\right| + \left|{m}\right| \cdot \left|{x_n - l}\right|          Modulus of Product          
\le K \cdot \left|{y_n - m}\right| + \left|{m}\right| \cdot \left|{x_n - l}\right|                    
= zn                    


But x_n \to l as n \to \infty.

So \left|{x_n - l}\right| \to 0 as n \to \infty from Convergent Sequence Minus Limit.

Similarly \left|{y_n - m}\right| \to 0 as n \to \infty.

From the Combined Sum Rule above, \lim_{n \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m, z_n \to 0 as n \to \infty.

The result follows by the Squeeze Theorem for Sequences of Complex Numbers (which applies as well to real as to complex sequences).

\blacksquare


[edit] Proof of Quotient Rule

We need to show that \lim_{n \to \infty} \frac {x_n} {y_n} = \frac l m, provided that m \ne 0.


As y_n \to m as n \to \infty, it follows from Modulus of Limit that \left|{y_n}\right| \to \left|{m}\right| as n \to \infty.

As m \ne 0, it follows from the definition of the modulus of m that \left|{m}\right| > 0.

From Sequence Converges to Within Half Limit, we have \exists N: \forall n > N: \left|{y_n}\right| > \frac {\left|{m}\right|} 2.

Now, for n > N, consider:

\left|{\frac {x_n} {y_n} - \frac {l} {m}}\right| = \left|{\frac {m x_n - y_n l} {m y_n}}\right|                    
< \frac 2 {\left|{m}\right|^2} \left|{m x_n - y_n l}\right|                    

By the above, m x_n - y_n l \to ml - ml = 0 as n \to \infty.

The result follows by the Squeeze Theorem for Sequences of Complex Numbers (which applies as well to real as to complex sequences).

\blacksquare


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