Can I just separate them?

Calculus Level 2

Find:. ( 1 1 2 + 1 3 2 + 1 5 2 + … ) ÷ ( 1 2 2 + 1 4 2 + 1 6 2 + … ) \large \left( \frac1{1^2} + \frac1{3^2} + \frac1{5^2} + \ldots \right) \div \left( \frac1{2^2} + \frac1{4^2} + \frac1{6^2} + \ldots \right)

Given that ∑ m = 1 ∞ 1 m 2 = 1 + 1 2 2 + 1 3 2 + … = π 2 6 \displaystyle \sum_{m=1}^\infty \frac1{m^2} = 1 + \frac1{2^2} + \frac1{3^2} + \ldots = \frac{\pi^2}6 .. Please reshare if you like this problem. Try this also .

5/3 3 4/3 1 3/2 2

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Gaurav Jain by Gaurav Jain , 22, India

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3 solutions

Mainak Pal
Jun 13, 2015

You don't need any values to evaluate it. [ 1 + 1 3 2 + 1 5 2 + . . . ∞ ] = [ 1 + 1 2 2 + 1 3 2 + 1 4 2 + 1 5 2 + . . . ∞ ] − [ 1 2 2 + 1 4 2 + 1 6 2 + . . . ∞ ] a n d , [ 1 2 2 + 1 4 2 + 1 6 2 + . . . ∞ ] = 1 2 2 [ 1 + 1 2 2 + 1 3 2 + 1 4 2 + 1 5 2 + . . . ∞ ] S o , [ 1 + 1 3 2 + 1 5 2 + . . . ∞ ] ÷ [ 1 2 2 + 1 4 2 + 1 6 2 + . . . ∞ ] = [ 1 + 1 2 2 + 1 3 2 + 1 4 2 + 1 5 2 + . . . ∞ ] [ 1 2 2 + 1 4 2 + 1 6 2 + 1 8 2 . . . ∞ ] − 1 = [ 1 + 1 2 2 + 1 3 2 + 1 4 2 + 1 5 2 + . . . ∞ ] 1 2 2 [ 1 + 1 2 2 + 1 3 2 + 1 4 2 + 1 5 2 + . . . ∞ ] − 1 = 3 \left[ 1+\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...\infty \right] =\left[ 1+\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...\infty \right] -\left[ \frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\frac { 1 }{ { 6 }^{ 2 } } +...\infty \right] \\ \\ and,\quad \left[ \frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\frac { 1 }{ { 6 }^{ 2 } } +...\infty \right] =\frac { 1 }{ { 2 }^{ 2 } } \left[ 1+\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...\infty \right] \\ \\ So,\\ \left[ 1+\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...\infty \right] \div \left[ \frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\frac { 1 }{ { 6 }^{ 2 } } +...\infty \right] =\frac { \left[ 1+\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...\infty \right] }{ \left[ \frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\frac { 1 }{ { 6 }^{ 2 } } +\frac { 1 }{ { 8 }^{ 2 } } ...\infty \right] } -1\\ =\frac { \left[ 1+\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...\infty \right] }{ \frac { 1 }{ { 2 }^{ 2 } } \left[ 1+\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 4 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...\infty \right] } -1=3

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good solution !

Gaurav Jain - 5 years, 12 months ago
Rwit Panda
Jun 15, 2015

In the denominator, take 1/4 common and you get an inverse square series. So the value of the denominator is pi^2/24. Another observation is that the numerator is an (inverse square series) - (denominator). So it has a value of 3(pi^2)/24. The fraction thus holds a value of 3.

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Gaurav Jain
Jun 12, 2015

The numerator one has the value (pi^2)/8 and the denominator has (pi^2)/24. The Lord Brounker's formula.

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Moderator note:

This is not a complete solution. Can you derive the values of π 2 8 \frac{\pi^2}{8} and π 2 24 \frac{\pi^2}{24} yourself?

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