Let $$a_n = \frac{1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}.$$ (a) Prove that $\lim_{n\to \infty}a_n$ exists. (b) Show that $$a_n = \frac{\left(1-\frac1{2^2}\right)\left(1-\frac1{4^2}\right)\left(1-\frac1{6^2}\right)\cdots\left(1-\frac{1}{(2n)^2}\right)}{(2n+1)a_n}.$$ (c) Find $\lim_{n\to\infty}a_n$ and justify your answer

[b]2.[/b] Show that the series $\sum_{n=1}^{\infty}\frac{1}{n}sin(asin(\frac{2n\pi}{N}))e^{bcos(\frac{2n\pi}{N})}$ is convergent for every positive integer N and any real numbers a and b. [b](S. 25)[/b]

Define the numbers $a_0, a_1, \ldots, a_n$ in the following way: \[ a_0 = \frac{1}{2}, \quad a_{k+1} = a_k + \frac{a^2_k}{n} \quad (n > 1, k = 0,1, \ldots, n-1). \] Prove that \[ 1 - \frac{1}{n} < a_n < 1.\]

Define the sequences $ a_n, b_n, c_n$ as follows. $ a_0 \equal{} k, b_0 \equal{} 4, c_0 \equal{} 1$. If $ a_n$ is even then $ a_{n \plus{} 1} \equal{} \frac {a_n}{2}$, $ b_{n \plus{} 1} \equal{} 2b_n$, $ c_{n \plus{} 1} \equal{} c_n$. If $ a_n$ is odd, then $ a_{n \plus{} 1} \equal{} a_n \minus{} \frac {b_n}{2} \minus{} c_n$, $ b_{n \plus{} 1} \equal{} b_n$, $ c_{n \plus{} 1} \equal{} b_n \plus{} c_n$. Find the number of positive integers $ k < 1995$ such that some $ a_n \equal{} 0$.

A sequence of real numbers is called a [i]Fibonacci [/i] sequence if $$t_{n+2} = t_{n+1} + t_n$$ for $n= 1,2,3,. .$ . Two Fibonacci sequences are said to be [i]essentially different[/i] if the terms of one sequence cannot be obtained by multiplying the terms of the other by a constant. For example, the Fibonacci sequences $1,2,3,5,8,...$ and $1,3,4,7,11,...$ are essentially different, but the sequences $1,2,3,5,8,...$ and $2,4,6,10,16,...$ are not. (a) Prove that there exist real numbers $p$ and $q$ such that the sequences $1,p,p^2,p^3,...$ and $1,q,q^2,q^3,...$ are essentially different Fibonacci sequences. (b) Let $a_1,a_2,a_3,...$ and $b_1,b_2,b_3,...$ be essentially different Fibonacci sequences. Prove that for every Fibonacci sequence $t_1,t_2,t_3,...$, there exists exactly one number $\alpha$ and exactly one number $\beta$, such that: $$t_n = \alpha a_n + \beta b_n$$ for $n = 1,2,3,...$ (c) $t_1,t_2,t_3,...$, is the Fibonacci sequence with $t_1 = 1$ and $t_2= 2$. Express $t_n$ in terms of $n$.

Let $m, n$ be positive integers. Consider a sequence of positive integers $a_1, a_2, ... , a_n$ that satisfies $m = a_1 \ge a_2\ge ... \ge a_n \ge 1$. Then define, for $1\le  i\le  m$, $b_i =$ # $\{ j \in \{1, 2, ... , n\}: a_j \ge i\}$, so $b_i$ is the number of terms $a_j $ of the given sequence for which $a_j  \ge i$. Similarly, we define, for $1\le   j \le  n$, $c_j=$ # $\{ i \in \{1, 2, ... , m\}: b_i \ge j\}$ , thus $c_j$ is the number of terms bi in the given sequence for which $b_i \ge j$. E.g.: If $a$ is the sequence $5, 3, 3, 2, 1, 1$ then $b$ is the sequence $6, 4, 3, 1, 1$. (a) Prove that $a_j = c_j $ for $1  \le j  \le n$. (b) Prove that for $1\le  k \le m$: $\sum_{i=1}^{k} b_i = k \cdot b_k + \sum_{j=b_{k+1}}^{n} a_j$.

Find all positive integers $n$ such that there exists a sequence of positive integers $a_1$, $a_2$,$\ldots$, $a_n$ satisfying: \[a_{k+1}=\frac{a_k^2+1}{a_{k-1}+1}-1\] for every $k$ with $2\leq k\leq n-1$. [i]Proposed by North Korea[/i]

Let $\{x_n\}$ be a sequence of real numbers in the interval $[0,1)$. Prove that there exists a sequence $1 < n_1 < n_2 < n_3 < \cdots$ of positive integers such that the following limit exists $$\lim_{i,j \to \infty} x_{n_i+n_j}. $$ That is, there exists a real number $L$ such that for every $\epsilon > 0,$ there exists a positive integer $N$ such that if $i,j > N$, then $|x_{n_i+n_j}-L| < \epsilon.$

Let $a_1 \le a_2 \le ... \le a_n$ be real numbers for which $$\sum_{i=1}^{n} a_i^{2k+1} = 0$$ holds for all integers $0 \le k < n$. Show that in this case, $a_i = -a_{n+1-i}$ holds for all $1 \le i \le n$.

Let $a_1,a_2,a_3,\dots$ be an infinite sequence of non-zero reals satisfying \[a_{i} = \frac{a_{i-1}a_{i-2}-2}{a_{i-3}}\]for all $i\geq 4$. Determine all positive integers $n$ such that if $a_1,a_2,\dots,a_n$ are integers, then all elements of the sequence are integers.

The sequence $a_0, a_1, a_2, ..$ is defined by $a_0 = a, a_{n+1} = a_n + L(a_n)$, where $L(m)$ is the last digit of $m$ (eg $L(14) = 4$). Suppose that the sequence is strictly increasing. Show that infinitely many terms must be divisible by $d = 3$. For what other d is this true?

The sequence $(a_n)_{n\geq1}$ is defined as: $$a_1=2, a_2=20, a_3=56, a_{n+3}=7a_{n+2}-11a_{n+1}+5a_n-3\cdot2^n.$$ Prove that $a_n$ is positive for every positive integer $n{}$. Find the remainder of the divison of $a_{673}$ to $673$.

Let $a_0, a_1, \ldots , a_n$ and $b_0, b_1, \ldots , b_n$ be sequences of real numbers such that $a_0 = b_0 \geqslant 0$, $a_n = b_n > 0$ and \[a_i=\sqrt{\frac{a_{i+1}+a_{i-1}}{2}},\quad b_i=\sqrt{\frac{b_{i+1}+b_{i-1}}{2}},\]for all $i=1,\ldots,n-1$. Prove that $a_1 = b_1$.

Does there exist an increasing sequence of positive integers for which any large enough integer can be expressed uniquely as the sum of two (possibly equal) terms of the sequence? [i]Proposed by Vlad Spătaru and David Anghel[/i]

Let ${a_n}$ be a non-increasing sequence of positive numbers. Prove that if for $n \ge 2001$, $na_{n} \le 1$, then for any positive integer $m \ge 2001$ and $x \in \mathbb{R}$, the following inequality holds: $\left | \sum_{k=2001}^{m} a_{k} \sin kx \right | \le 1 + \pi$

The sequence $x_1,x_2, ...$ is defined by the following equations: $$x_1=19, \ \ x_2=97, \ \ x_{n+2} =x_n - \frac{1}{x_{n+1}}$$ for $n \ge 1$. Prove that there exists a positive integer $k$ such that $x_k=0$ and find $k$. (A Berzinsh)

A finite sequence of natural numbers $a_1, a_2, \dots, a_n$ is given. A sub-sequence $a_{k+1}, a_{k+2}, \dots, a_l$ will be called a [i]repetition[/i] if there exists a natural number $p\leq \frac{l-k}2$ such that $a_i=a_{i+p}$ for $k+1\leq i\leq l-p$, but $a_i\neq a_{i+p}$ for $i=k$ (if $k>0$) and $i=l-p+1$ (if $l<n$). Show that the sequence contains less than $n$ repetitions.

The sequences $(x_n)$ and $(y_n)$ are defined as follows: $$ x_{n+1} = \frac{x_n+2}{x_n+1},\quad y_{n+1}=\frac{y_n^2+2}{2y_n} \quad \text{ for } n= 0,1,2,\ldots.$$ Prove that for every integer $ n\geq 0 $ the equality $ y_n = x_{2^n-1} $ holds.

Let $k$ be a positive integer. A sequence of integers $a_1, a_2, \cdots$ is called $k$-pop if the following holds: for every $n \in \mathbb{N}$, $a_n$ is equal to the number of distinct elements in the set $\{a_1, \cdots , a_{n+k} \}$. Determine, as a function of $k$, how many $k$-pop sequences there are. [i]Proposed by Sutanay Bhattacharya[/i]

Let $L_1=1$, $L_2=3$, and $L_{n+2}=L_{n+1}+L_n$ for $n\geq1$. How many terms in the sequence $L_1, L_2, L_3, \dots, L_{2023}$ are even? $\textbf{(A) }673\qquad\textbf{(B) }1011\qquad\textbf{(C) }675\qquad\textbf{(D) }1010\qquad\textbf{(E) }674$

Positive integers $a_1, a_2, \dots, a_{2023}$ satisfy the following conditions. [list] [*] $a_1 = 5, a_2 = 25$ [*] $a_{n + 2} = 7a_{n+1}-a_n-6$ for each $n = 1, 2, \dots, 2021$ [/list] Prove that there exist integers $x, y$ such that $a_{2023} = x^2 + y^2.$

Let $n$ be a positive integer. Show that there is a prime $p$ and a sequence $\left(a_k\right)_{k\ge1}$ of positive integers such that the sequence $\left(p+na_k\right)_{k\ge1}$ consists of distinct primes.

The sequence of real numbers $a_0,a_1,a_2,\ldots$ is defined recursively by \[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]Show that $ a_{n} > 0$ for all $ n\geq 1$. [i]Proposed by Mariusz Skalba, Poland[/i]

A sequence $x_0, x_1, x_2, . . .$ is given by $x_0 = 8$ and $x_{n+1} =\frac{1 + x_n}{1- x_n}$ for $n = 0, 1, 2, . . . .$ Determine the number $x_{2013}$.

Let $0 < \frac{1}{q} \leq \frac{1}{p} < 1$ and $\frac{1}{p}+\frac{1}{q}=1$. Let $u_k$, $v_k$, $a_k$ and $b_k$ be non-negative real sequences such as $u^2_k > a^p_k$ and $v_k > b^q_k$, where $k = 1, 2,\cdots , n$. If $0 < m_1\leq u_k \leq M_1$ and $0 < m_2 \leq v_k \leq M_2$ , then $$\left(\sum \limits_{k=1}^n\left(l^p\left(u_k+v_k\right)^2-\left(a_k+b_k\right)^p\right)\right)^{\frac{1}{p}}\geq \left(\sum \limits_{k=1}^n\left(u_k^2-a_k^p\right)\right)^{\frac{1}{p}}\left(\sum \limits_{k=1}^n\left(v_k^2-b_k^p\right)\right)^{\frac{1}{p}}$$where $$l=\frac{M_1M_2+m_1m_2}{2\sqrt{m_1M_1m_2M_2}}$$