Basel Problem Extension

Algebra Level 3

In 1734, Euler solved the Basel problem, proving that: π 2 6 = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ \frac{\pi^2}{6} = \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots

Mary accidentally burns a lot of the fractions in the series, so now only the fractions with odd denominators remain. So, the series is now,

π 2 N = 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + ⋯ \frac{\pi^2}{N}= \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots

for some integer N N . Compute N N .


The answer is 8.

Bryan Hung by Bryan Hung , 20, USA

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2 solutions

Bryan Hung
Dec 21, 2017

Let's instead find the sum of all the fractions in the series with an even denominator. Note that:

1 2 2 + 1 4 2 + 1 6 2 + 1 8 2 + ⋯ = 1 4 ( 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ ) = 1 4 ( π 2 6 ) = π 2 24 \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\cdots = \frac{1}{4}{(}\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots{)}=\frac{1}{4}{(}\frac{\pi^2}{6}{)}=\frac{\pi^2}{24}

Now it's easy:

( 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + ⋯ ) + ( 1 2 2 + 1 4 2 + 1 6 2 + 1 8 2 + ⋯ ) = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ {(}\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots{)} + {(}\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\cdots{)} =\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots

( 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + ⋯ ) + π 2 24 = π 2 6 {(}\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots{)} + \frac{\pi^2}{24} =\frac{\pi^2}{6}

1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + ⋯ = π 2 8 \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots =\boxed{\frac{\pi^2}{8}}

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Chew-Seong Cheong
Dec 21, 2017

S = 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + ⋯ = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ − ( 1 2 2 + 1 4 2 + 1 6 2 + 1 8 2 + ⋯ ) = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ − 1 4 ( 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ ) = π 2 6 − 1 4 ( π 2 6 ) = π 2 8 \begin{aligned} S & = \frac 1{1^2} + \frac 1{3^2} + \frac 1{5^2} + \frac 1{7^2} + \cdots \\ & = \frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + \cdots - \left( \frac 1{2^2} + \frac 1{4^2} + \frac 1{6^2} + \frac 1{8^2} + \cdots \right) \\ & = {\color{#3D99F6}\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + \cdots} - \frac 14 \left({\color{#3D99F6}\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + \cdots} \right) \\ & = {\color{#3D99F6} \frac {\pi^2}6} - \frac 14 \left({\color{#3D99F6} \frac {\pi^2}6} \right) \\ & = \frac {\pi^2}8 \end{aligned}

⟹ N = 8 \implies N = \boxed{8}

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