Let us write \([n]=\{1,2,\ldots,n\}\). Let \(a_n\) be the number of all functions \(f:[n]\to [n]\) such that \(f([n])=[k]\) for some positive integer \(k\). Prove that \[a_n=\sum_{k=0}^{\infty} \frac{k^n}{2^{k+1}}.\]

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Let us write \([n]=\{1,2,\ldots,n\}\). Let \(a_n\) be the number of all functions \(f:[n]\to [n]\) such that \(f([n])=[k]\) for some positive integer \(k\). Prove that \[a_n=\sum_{k=0}^{\infty} \frac{k^n}{2^{k+1}}.\]

Prove that for all positive integers m and n, there is a positive integer k such that \[ (\sqrt{m}+\sqrt{m-1})^n = \sqrt{k}+\sqrt{k-1}.\]

The best solution was submitted by Jesek Lee (이재석), 수리과학과 2007학번. Congratulations!

Here is his Solution of Problem 2011-2.

Alternative solutions were submitted by 김인환 (2010학번, +3), 박민재 (2011학번, +3), 김지원 (2010학번, +3),강동엽 (전산학과 2009학번, +3), 서기원 (수리과학과 2009학번, +3), 김재훈 (EEWS대학원 2010학번, +3), 김현수 (한국과학영재학교 3학년, +3), 어수강 (홍익대학교 수학교육학과 2004학번, +3).

Evaluate the sum \[ \sum_{n=1}^{\infty} \frac{n \sin n}{1+n^2}. \]

The best solution was submitted by Minjae Park (박민재), 2011학번. Congratulations!

Here is his Solution of Problem 2011-1.

A solution for an incorrect version of the problem was submitted by 구도완 (해운대고등학교 3학년).

Prove that for all positive integers m and n, there is a positive integer k such that \[ (\sqrt{m}+\sqrt{m-1})^n = \sqrt{k}+\sqrt{k-1}.\]

Evaluate the sum \[ \sum_{n=1}^{\infty} \frac{n \sin n}{1+n^2}. \]

(UPDATED: 2011.2.18) I have fixed a typo in the formula. Initially the following formula \[ \sum_{n=1}^{\infty} \frac{\sin n}{1+n^2}\] was posted but it does not seem to have a closed form answer. I’m sincerely sorry!