Problem 12067 - 08 - P. Bracken (USA).

对于正整数$n$.令$$\beta_n=6n+12n^2(\gamma-\gamma_n),$$
其中$\gamma_n=H_n-\ln n$, $\displaystyle H_n=\sum_{j=1}^n\frac{1}{j}$为第$n$个调和数(Harmonic number),且$\gamma$为Euler常数.证明:对所有$n$,均有$\beta_{n+1}>\beta_n$.

Problem 12066 - 08 - Xiang-Qian Chang (USA).
设$n$和$k$是大于$1$的正整数, $A$是$n\times n$正定Hermite矩阵.证明
$$\left( \det A \right) ^{1/n}\le \left( \frac{\mathrm{tr}^k\left( A \right) -\mathrm{tr}\left( A^k \right)}{n^k-n} \right) ^{1/k}.$$
Problem 12064 - 08 - C. A. Hernandez Melo (Brazil).
设$f$为凸的,从$[1,\infty)$到$\mathbb{R}$的连续可微函数,使得对所有$x\geq 1$,均有$f'(x)>0$.证明反常积分$$\int_1^\infty\frac{dx}{f'(x)}$$收敛当且仅当级数$$\sum_{n=1}^\infty\left(f^{-1}(f(n)+\varepsilon)-n\right)$$对所有$\varepsilon>0$均收敛.
Problem 12063 - 08 - H. Ohtsuka (Japan).
设$p$和$q$为实数,且$p>0,q>-p^2/4$.令$U_0=0,U_1=1$,而且对于$n\geq 0$,有$U_{n+2}=pU_{n+1}+qU_n$.计算
$$\lim_{n\rightarrow \infty}\sqrt{U_{1}^{2}+\sqrt{U_{2}^{2}+\sqrt{U_{4}^{2}+\sqrt{\cdots +\sqrt{U_{2^{n-1}}^{2}}}}}}.$$
Problem 12060 - 07 - O. Furdui and A. Sintamarian (Romania).
证明$$\sum_{n=2}^{\infty}{\frac{H_nH_{n+1}}{n^3-n}}=\frac{5}{2}-\frac{\pi ^2}{24}-\zeta \left( 3 \right),$$
其中$\displaystyle H_n=\sum_{j=1}^n\frac{1}{j}$为第$n$个调和数(Harmonic number).
Problem 12057 - 07 - P. Korus (Hungary).
(a)设$a_1=1,a_2=2$且对任意正整数$k$,有
$$a_{2k+1}=\frac{a_{2k-1}+a_{2k}}{2},\quad a_{2k+2}=\sqrt{a_{2k}a_{2k+1}}.$$
求数列$\{a_n\}$的极限.
(b)设$b_1=1,b_2=2$且对任意正整数$k$,有
$$b_{2k+1}=\frac{b_{2k-1}+b_{2k}}{2},\quad b_{2k+2}=\frac{2b_{2k}b_{2k+1}}{b_{2k}+b_{2k+1}}.$$
求数列$\{b_n\}$的极限.
Problem 12054 - 06 - C. I. Valean (Romania).
证明$$\int_0^1{\frac{\arctan x}{x}\ln \frac{1+x^2}{\left( 1-x \right) ^2}dx}=\frac{\pi ^3}{16}.$$
Problem 12051 - 06 - P. Ribeiro (Portugal).
证明$$
\sum_{n=0}^{\infty}{\left( \begin{array}{c}
2n\\
n\\
\end{array} \right) \frac{1}{4^n\left( 2n+1 \right) ^3}}=\frac{\pi ^3}{48}+\frac{\pi \ln ^22}{4}.
$$
Problem 12049 - 06 - Z. K. Silagadze (Russia).
对所有满足$m\leq n$的非负整数$m$和$n$.证明
$$
\sum_{k=m}^n{\frac{\left( -1 \right) ^{k+m}}{2k+1}\left( \begin{array}{c}
n+k\\
n-k\\
\end{array} \right) \left( \begin{array}{c}
2k\\
k-m\\
\end{array} \right)}=\frac{1}{2n+1}.
$$