Problem B. 4019.

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B. 4019. Prove that

$\frac{1}{3^2} + \frac{1}{5^2} + \dots + \frac{1}{{(2n+1)}^2} < \frac{1}{4}.$

for every positive integer n.

(Competition problem from the Highlands)

(4 points)

Deadline expired.

Sorry, the solution is published in Hungarian only.

Megoldás: Minden k pozitív egész számra

$\Bigl(\frac{1}{2k+1}\Bigr)^2 <\frac{1}{4k^2+4k}=\frac{1}{4}\cdot\frac{1}{k(k+1)} =\frac{1}{4}\Bigl(\frac{1}{k}-\frac{1}{k+1}\Bigr),$

ezért a szóban forgó összeg kisebb, mint

$\frac{1}{4}\Bigl\{\Bigl(\frac{1}{1}-\frac{1}{2}\Bigr)+ \Bigl(\frac{1}{2}-\frac{1}{3}\Bigr)+\ldots+ \Bigl(\frac{1}{n}-\frac{1}{n+1}\Bigr)\Bigr\}= \frac{1}{4}\Bigl(1-\frac{1}{n+1}\Bigr)<\frac{1}{4}.$

Statistics on problem B. 4019.

178 students sent a solution.

4 points: 133 students.

3 points: 8 students.

2 points: 5 students.

1 point: 6 students.

0 point: 26 students.

Problems in Mathematics of KöMaL, September 2007

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