Limit Constant Multiple/Power Laws for Convergent Sequences

# Limit Constant Multiple/Power Laws for Convergent Sequences

We will now proceed to specifically look at the limit constant multiple and power laws (law 5 and law 6 from the Limit of a Sequence page) and prove their validity.

## Constant Multiple Law for Convergent Sequences

Law 5 (Constant Multiple Law of Convergent Sequences): If the limit of the sequence $\{ a_n \}$ is convergent, that is $\lim_{n \to \infty} a_n = A$, and $k$ is a constant, then $\lim_{n \to \infty} ka_n = k \lim_{n \to \infty} a_n = kA$. |

**Proof of Law 5:**

- Let $k$ be a constant such that $k \neq 0$. We note that the case that $k = 0$ is trivial since then $\lim_{n\ \to \infty} 0a_n = \lim_{n \to \infty} 0 = 0$. Let $\{ a_n \}$ be a convergent sequence so that $\lim_{n \to \infty} a_n = A$ which implies that $\forall \epsilon \: \exists N_1 \in \mathbb{N}$ such that if $n ≥ N_1$ then $\mid a_n - A \mid < \frac{\epsilon}{\mid k \mid}$.

- Now let $\epsilon > 0$ be given. We want to show $\lim_{n \to \infty} ka_n = kA$, that is, $\forall \epsilon > 0 \: \exists N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid k_an - kA \mid < \epsilon$. Working with this inequality:

\begin{align} \quad \mid k a_n - kA \mid = \mid k(a_n - A) \mid = \mid k \mid \mid a_n - A \mid < \epsilon \end{align}

- We choose $N = N_1$ and therefore if $n ≥ N$ then $\mid a_n - A \mid < \frac{\epsilon}{\mid k \mid}$ and so $\lim_{n \to \infty} ka_n = kA$. $\blacksquare$

## Power Law for Convergent Sequences

Law 6 (Power Law of Convergent Sequences): If the limit of the sequence $\{ a_n \}$ is convergent, that is $\lim_{n \to \infty} a_n = A$ and $k$ is a non-negative integer, then $\lim_{n \to \infty} (a_n)^k = \left ( \lim_{n \to \infty} a_n \right )^k = (A)^k$ provided that $k \in \mathbb{N}$. |

**Proof of Law 6:**Recall that from the power law for sequences that if $\{ a_n \}$ and $\{ b_n \}$ are convergent sequences such that $\lim_{n \to \infty} a_n = A$ and $\lim_{n \to \infty} b_n = B$, then $\lim_{n \to \infty} [a_n b_n] = AB$. The power law is just a special case of this.

- Consider $\{ b_n \} = \{ a_n \}$ and therefore by the power law $\lim_{n \to \infty} [a_n a_n] = \lim_{n \to \infty} (a_n)^2 = AA = A^2$. Therefore $\{ a_n^2 \}$ is a convergent sequence. Now continue to the process by letting $\{b_n \} = \{ a_n \}^2$ and therefore $\lim_{n \to \infty} a_n a_n^2 = AA^2 = A^3$, etc… so we deduce if $\{ a_n \}$ is convergent then $\lim_{n \to \infty} (a_n)^k = A^k$.