User:Dezhidki/Sandbox

Theorem
Let $(a_n)$ be any sequence and $(b_n)$ be a sequence of positive real numbers so that $\displaystyle \sum_{i \mathop = 0}^{\infty}b_n = \infty$. If
 * $\lim\limits_{n \to \infty }\dfrac{a_n}{b_n} = L \in \R$

then also
 * $\lim\limits_{n \to \infty }\dfrac{a_1 + a_2+ \cdots + a_n}{b_1 + b_2 + \cdots + b_n} = L \in \R$.

Proof
Define the following sums:
 * $\displaystyle A_n = \sum_{i \mathop = 1}^{n}a_i$ and
 * $\displaystyle B_n = \sum_{i \mathop = 1}^{n}b_i$.

Let $\varepsilon > 0$ and $\mu = \dfrac{\varepsilon}{2}$. By the definition of limits of sequences, there exists $k \in \N$ such that
 * $\displaystyle \left( L - \mu \right)b_n < a_n < \left( L + \mu \right)b_n$ for all $n > k$.

By summing terms $a_1, \ldots, a_k, \ldots, a_n$ we obtain

Dividing all sides by $B_n$ we get
 * $\displaystyle \frac{A_k + (L-\mu)B_k}{B_n} + (L - \mu) < \frac{A_n}{B_n} < (L + \mu) + \frac{A_k + (L+\mu)B_k}{B_n}$.

Let $k$ be fixed. Since $B_n \to \infty$ as $n \to \infty$, sequence $\displaystyle \left( \frac{A_k + (L \pm \varepsilon)B_k}{B_n} \right)$ must converge to zero. Thus there exists $N > k > 0$ such that
 * $\displaystyle \left| \frac{A_k + (L \pm \mu)B_k}{B_n} \right| < \mu$ for all $n > N$.

Thus, by substituting, obtain

which by definition of limit of sequences implies that $\displaystyle \lim\limits_{n \to \infty}\frac{A_n}{B_n} = L$ as needed.

Corollary
Let $(a_n)$ and $(b_n)$ be sequences such that $\lim\limits_{n \to \infty}b_n = \infty$. If
 * $\displaystyle \lim\limits_{n \to \infty}\frac{a_n - a_{n-1}}{b_n - b_{n-1}} = L \in \R$

then also
 * $\displaystyle \lim\limits_{n \to \infty}\frac{a_n}{b_n} = L$.

Proof
Define the following sequences:
 * $x_n = a_n - a_{n-1}$, $x_1 = a_1$ and
 * $y_n = b_n - b_{n-1}$, $y_1 = b_1$.

It follows that $\displaystyle \sum_{i \mathop = 1}^{n}x_i = a_n$, $\displaystyle \sum_{i \mathop = 1}^{n}y_i = b_n$ and
 * $\displaystyle \lim\limits_{n \to \infty}\frac{a_n - a_{n-1}}{b_n - b_{n-1}} = \lim\limits_{n \to \infty}\frac{x_n}{y_n} =  L $

Since $b_n \to \infty$ as $n \to \infty$, there exists $N \in \N$ such that
 * $b_n > 0$ for all $n > N$.

Thus it follows from the main theorem that
 * $\displaystyle \lim\limits_{n \to \infty}\frac{\sum_{i \mathop = 1}^{n}x_i}{\sum_{i \mathop = 1}^{n}y_i} = \lim\limits_{n \to \infty}\frac{a_n}{b_n} = L$.