0

Could someone help me to prove this theorem. Thank you

Theorem Suppose that the sequence of complex number $ \{a_n\}_{n\geq 1} $ converges. Then the sequence of its averages also converges, and their limits coincide; that is \begin{align*} \lim\limits_{n\to\infty}\frac{a_1+a_2+\cdots+a_n}{n}=\lim\limits_{n\to\infty}a_n. \end{align*}

E-moon
  • 19
  • I find it helpful to see if it feels right. The first $a$s can do anything they want, but after a while they are all close to the limit. That suggests that you just need $n$ large enough to dilute the errors contributed by the first terms. – Ross Millikan Mar 02 '20 at 04:23

1 Answers1

1

Let $\displaystyle \lim_{n \to \infty} a_n = L$ and because convergent sequences are bounded, let say $|a_n| < M$.

$\displaystyle \lim_{n \to \infty} a_n = L$ means $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ such that $n>N \implies |a_n - L| < \epsilon$.

We need to show, given $\epsilon_1 > 0$ we can find $N_1 \in \mathbb{N}$ such that $\displaystyle n>N_1 \implies \left| \frac{a_1 + a_2 + \cdots + a_n}{n} - L \right| < \epsilon$.

Now let we are given $\epsilon_1 > 0$.

Choosing $\epsilon = \epsilon_1/2$ in the second line we have $N_{\epsilon} \in \mathbb{N}$ such that $n>N_{\epsilon} \implies |a_n - L| < \epsilon_1/2$. Thus,

$\displaystyle \left| \frac{a_1 + a_2 + \cdots + a_n}{n} - L \right| = \left| \frac{a_1 + a_2 + \cdots + a_n - nL}{n} \right| = \left| \frac{(a_1 - L) + (a_2 - L) + \cdots + (a_n - L)}{n} \right| $

$\displaystyle = \left| \frac{(a_1 - L) + (a_2 - L) + \cdots + (a_{N_{\epsilon}} - L) + (a_{N_{\epsilon} + 1} - L) + (a_{N_{\epsilon} + 2} - L) + \cdots + (a_{n} - L)}{n} \right|$

$\displaystyle < \frac{|a_1 - L| + |a_2 - L| + \cdots + |a_{N_{\epsilon}} - L| + |a_{N_{\epsilon} + 1} - L| + |a_{N_{\epsilon} + 2} - L| + \cdots + |a_{n} - L|}{n}$

$\displaystyle < \frac{|a_1| + |L| +|a_2| + |L| + \cdots + |a_{N_{\epsilon}}| + |L| + \epsilon_1/2 + \epsilon_1/2 + \cdots + \epsilon_1/2}{n}$

$\displaystyle < \frac{N_{\epsilon}(M + |L|) + (n-N_{\epsilon}) \epsilon_1/2}{n} = \frac{N_{\epsilon}(M + |L|)}{n} + \frac{n-N_{\epsilon}}{n} \frac{\epsilon_1}{2} < \frac{N_{\epsilon}(M + |L|)}{n} + \frac{\epsilon_1}{2}$ for $n>N_{\epsilon}$.

Since $N_{\epsilon}$ is fixed and $M$, $L$ are constants, we can find $N_2 \in \mathbb{N}$ so that $\displaystyle \frac{N_{\epsilon}(M + |L|)}{n} < \frac{\epsilon_1}{2}$ for $n>N_2$ because $\displaystyle \frac{N_{\epsilon}(M + |L|)}{n} \to 0$ as $n \to \infty$.

Finally, choosing $N_1 = \max\{ N_{\epsilon}, N_2 \}$ in the third line we get

$\displaystyle n>N_1 \implies \left| \frac{a_1 + a_2 + \cdots + a_n}{n} - L \right| < \frac{N_{\epsilon}(M + |L|)}{n} + \frac{\epsilon_1}{2} < \frac{\epsilon_1}{2} + \frac{\epsilon_1}{2} = \epsilon_1.$ Hence we prove that

for a given $\epsilon_1 > 0$ we are managed to find $N_1 \in \mathbb{N}$ so that the statement in the third line holds true.

Can Turkun
  • 424