Plus minus plus minus plus minus

Calculus Level 2

1 , 1 2 2 , 1 3 2 , 1 4 2 , … 1, \frac1{2^2}, \frac1{3^2} , \frac1{4^2} , \ldots

The sum of all the numbers above gives a value of A A .
Whereas the alternating sum of the numbers above gives a value of B B .

What is the ratio between A ÷ B A \div B ?


The answer is 2.

Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

We have that the alternating sum B = 1 − 1 2 2 + 1 3 2 − 1 4 2 + 1 5 2 − 1 6 2 + . . . . . . = B = 1 - \dfrac{1}{2^{2}} + \dfrac{1}{3^{2}} - \dfrac{1}{4^{2}} + \dfrac{1}{5^{2}} - \dfrac{1}{6^{2}} + ...... =

A − 2 ( 1 2 2 + 1 4 2 + 1 6 2 + . . . . . ) = A − 2 ( 1 2 2 + 1 ( 2 ∗ 2 ) 2 + 1 ( 2 ∗ 3 ) 2 + . . . . . ) = A - 2\left(\dfrac{1}{2^{2}} + \dfrac{1}{4^{2}} + \dfrac{1}{6^{2}} + ..... \right) = A - 2\left(\dfrac{1}{2^{2}} + \dfrac{1}{(2*2)^{2}} + \dfrac{1}{(2*3)^{2}} + ..... \right) =

A − 2 2 2 ( 1 + 1 2 2 + 1 3 2 + . . . . ) = A − 1 2 A = 1 2 A A - \dfrac{2}{2^{2}}\left(1 + \dfrac{1}{2^{2}} + \dfrac{1}{3^{2}} + .... \right) = A - \dfrac{1}{2}A = \dfrac{1}{2}A , and so A ÷ B = A ÷ ( A / 2 ) = 2 A \div B = A \div (A/2) = \boxed{2} .

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