Školjka

%V0 Suppose that $x_1, x_2, x_3, \ldots$ are positive real numbers for which $$x^n_n = \sum^{n-1}_{j=0} x^j_n$$ for $n = 1, 2, 3, \ldots$ Prove that $\forall n,$ $$2 - \frac{1}{2^{n-1}} \leq x_n < 2 - \frac{1}{2^n}.$$

%V0 Consider two monotonically decreasing sequences $\left( a_k\right)$ and $\left( b_k\right)$, where $k \geq 1$, and $a_k$ and $b_k$ are positive real numbers for every k. Now, define the sequences $c_k = \min \left( a_k, b_k \right)$; $A_k = a_1 + a_2 + ... + a_k$; $B_k = b_1 + b_2 + ... + b_k$; $C_k = c_1 + c_2 + ... + c_k$ for all natural numbers k. (a) Do there exist two monotonically decreasing sequences $\left( a_k\right)$ and $\left( b_k\right)$ of positive real numbers such that the sequences $\left( A_k\right)$ and $\left( B_k\right)$ are not bounded, while the sequence $\left( C_k\right)$ is bounded? (b) Does the answer to problem (a) change if we stipulate that the sequence $\left( b_k\right)$ must be $\displaystyle b_k = \frac {1}{k}$ for all k ?

%V0 Let $a_0$, $a_1$, $a_2$, ... be an infinite sequence of real numbers satisfying the equation $a_n=\left|a_{n+1}-a_{n+2}\right|$ for all $n\geq 0$, where $a_0$ and $a_1$ are two different positive reals. Can this sequence $a_0$, $a_1$, $a_2$, ... be bounded? Remark This one is from the IMO Shortlist 2004, but it's already published on the official BWM website und thus I take the freedom to post it here:

%V0 $a_{0},\ a_{1},\ a_{2},\dots$ is a sequence of real numbers such that $$a_{n + 1} = \left[a_{n}\right]\cdot \left\{a_{n}\right\}$$ prove that exist $j$ such that for every $i\geq j$ we have $a_{i + 2} = a_{i}$.

%V0 Let $a_{0}$, $a_{1}$, $a_{2}$, $...$ be a sequence of reals such that $a_{0} = - 1$ and $a_{n} + \frac {a_{n - 1}}{2} + \frac {a_{n - 2}}{3} + ... + \frac {a_{1}}{n} + \frac {a_{0}}{n + 1} = 0$ for all $n\geq 1$. Show that $a_{n} > 0$ for all $n\geq 1$.

%V0 Let $c > 2,$ and let $a(1), a(2), \ldots$ be a sequence of nonnegative real numbers such that $$a(m + n) \leq 2 \cdot a(m) + 2 \cdot a(n) \text{ for all } m,n \geq 1,$$ and $a\left(2^k \right) \leq \frac {1}{(k + 1)^c} \text{ for all } k \geq 0.$ Prove that the sequence $a(n)$ is bounded. Author: Vjekoslav Kovač, Croatia