Stolz–Cesàro theorem
Stolz–Cesàro theorem is a powerful tool for evaluating limits of sequences, and it is a discrete version of L'Hôpital's rule.
Theorem 1 (the \(\infty/\infty\) case):
Let \((a_n)_{n\in\mathbb N} \) and \((b_n)_{n\in\mathbb N} \) be two sequences of real numbers such that
(a) \(0<b_1 < b_2 < \cdots < b_n < \cdots \) and \( \displaystyle \lim_{n\to\infty} b_n = \infty \).
(b) \( \displaystyle \lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} = l \in \mathbb R \).Then, \( \displaystyle \lim_{n\to\infty} \dfrac{a_n}{b_n} \) exists and is equal to \( l\).
Theorem 2 (the \(0/0\) case):
Let \((a_n)_{n\in\mathbb N} \) and \((b_n)_{n\in\mathbb N} \) be two sequences of real numbers such that
(a) \( \displaystyle \lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = 0 \).
(b) \( (b_n) \) is strictly decreasing.
(c) \( \displaystyle \lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} = l \in \mathbb R \).Then, \( \displaystyle \lim_{n\to\infty} \dfrac{a_n}{b_n} \) exists and is equal to \( l\).
Theorem 3 (Reciprocal of Stolz–Cesàro lemma):
Let \((a_n)_{n\in\mathbb N} \) and \((b_n)_{n\in\mathbb N} \) be two sequences of real numbers such that
(a) \(0<b_1 < b_2 < \cdots < b_n < \cdots \) and \( \displaystyle \lim_{n\to\infty} b_n = \infty \).
(b) \( \displaystyle \lim_{n\to\infty} \dfrac{a_n}{b_n} = l \in \mathbb R \).
(c) \( \displaystyle \lim_{n\to\infty} \dfrac{b_n}{b_{n+1}} = L \in \mathbb R\backslash \{1\} \).Then, \( \displaystyle \lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} \) exists and is equal to \( l\).
Theorem 3 (Multiplicative property):
If the sequence \((x_n /x_{n-1} )_{n=1}^\infty \) has a limit, then \( \displaystyle \lim_{n\to\infty} \sqrt[n]{x_n} = \lim_{n\to\infty} \frac{x_n}{x_{n-1}} \).
\[\large \lim_{n\to \infty}\frac{1}{n^{2015}} \sum_{k=1}^{n}(k+2015)^{2014}\]
The previous limit is equal to a number of the form \(\dfrac{a}{b}\) where \(a\) and \(b\) are two coprime natural numbers. Find \(a+b.\)
\[\large{\lim_{n \to \infty} \dfrac{x_n \cdot x_{n+2}}{x_{n+1}^2} = 4}\]
Let \((x_n)_{n=1}^{\infty}\) be a sequence such that the above limit satisfies, and where \(x_i > 0\) for every \(i \geq 1\). Then find the value of \( \displaystyle {\lim_{n \to \infty} \sqrt[n^2]{x_n} }\).
\[S_{n}= \frac{n+1}{2^{n+1}}\sum_{i=1}^{n}\frac{2^{i}}{i}\]
Consider the sequence, find \( \displaystyle \lim_{n\rightarrow \infty }S_{n} \).
Let \(a_k\) denote the number of digits in \(2^k\). For instance, since \(2^4 = 16\), we have \(a_4 = 2\).
Compute the limit
\[\lim_{n\to\infty} \frac{1}{n^2} \sum_{k=1}^{n} a_k. \]
Compute \[ \large \lim_{ n\to{\infty}}{\dfrac{ 2016(1^{2015}+2^{2015}+ \cdots +n^{2015}) - n^{2016}}{2016(1^{2014}+2^{2014}+\cdots+n^{2014})}}.\]