Sequence and Series

Algebra Level 4

If a = ∑ r = 1 ∞ 1 r 2 \displaystyle a=\sum_{r=1}^\infty \dfrac1{r^2} and b = ∑ r = 1 ∞ 1 ( 2 r − 1 ) 2 \displaystyle b=\sum_{r=1}^\infty \dfrac1{(2r-1)^2} , find 3 a b \dfrac{3a}{b} .


The answer is 4.

Dharani Chinta by Dharani Chinta , 22, India

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2 solutions

a = ∑ r = 1 ∞ 1 r 2 = ζ ( 2 ) = π 2 6 ζ ( n ) is the Riemann zeta function. b = ∑ r = 1 ∞ 1 ( 2 r − 1 ) 2 = ∑ r = 1 ∞ 1 r 2 − ∑ r = 1 ∞ 1 ( 2 r ) 2 = ∑ r = 1 ∞ 1 r 2 − 1 4 ∑ r = 1 ∞ 1 r 2 = 3 4 ∑ r = 1 ∞ 1 r 2 = 3 4 ζ ( 2 ) = π 2 8 \begin{aligned} a & = \sum_{r=1}^\infty \frac1{r^2} = \color{#3D99F6}{\zeta (2)} = \frac{\pi^2}6\quad \quad \small \color{#3D99F6}{\zeta (n) \text{ is the Riemann zeta function.}} \\ b & = \sum_{r=1}^\infty \frac1{(2r-1)^2} \\ & = \sum_{r=1}^\infty \frac1{r^2} - \sum_{r=1}^\infty \frac1{(2r)^2} \\ & = \sum_{r=1}^\infty \frac1{r^2} - \frac14 \sum_{r=1}^\infty \frac1{r^2} \\ & = \frac34 \sum_{r=1}^\infty \frac1{r^2} = \frac34 \zeta(2) = \frac{\pi^2}8 \end{aligned}

⟹ 3 a b = 4 \implies \dfrac{3a}b = \boxed{4}

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Or maybe write ′ a ′ 'a' as:

a = ( 1 1 2 + 1 3 2 + 1 5 2 ⋯ ) ⏟ b + ( 1 2 2 + 1 4 2 + 1 6 2 + ⋯ ) ⏟ 1 2 2 ( a ) a=\underbrace{\left(\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}\cdots\right)}_{\large b}+\underbrace{\left(\dfrac1{2^2}+\dfrac{1}{4^2}+\dfrac1{6^2}+\cdots\right)}_{\dfrac{1}{2^2}(\large a)} ⟹ 3 a / 4 = b ⟹ 3 a / b = 4 \implies 3a/4=b\implies 3a/b=4

BTW (+1)..

Rishabh Jain - 5 years ago

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Good solution. Upvoted.

Chew-Seong Cheong - 5 years ago

@Dharani Chinta , we have been keying in your problems into LaTex for you. Please learn to key in your problems in LaTex. Use can use Toggle LaTex at the top left pull-down menu to see the LaTex. You can also place your mouse cursor on top the formulas to check the codes. Thanks.

Chew-Seong Cheong - 5 years ago

a = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + 1 5 2 + . . . b = 1 1 2 + 1 3 2 + 1 5 2 + . . . a − b = 1 2 2 + 1 4 2 + 1 6 2 + . . . = 1 4 ( 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + 1 5 2 + . . . ) = 1 4 a 3 4 a = b ⇒ a = 4 3 b ⇒ 3 a b = 4 a=\frac{1}{1}^2+\frac{1}{2}^2+\frac{1}{3}^2+\frac{1}{4}^2+\frac{1}{5}^2+...\\b=\frac{1}{1}^2+\frac{1}{3}^2+\frac{1}{5}^2+...\\a-b=\frac{1}{2}^2+\frac{1}{4}^2+\frac{1}{6}^2+...=\frac{1}{4}\left(\frac{1}{1}^2+\frac{1}{2}^2+\frac{1}{3}^2+\frac{1}{4}^2+\frac{1}{5}^2+...\right)=\frac{1}{4}a\\\frac{3}{4}a=b\Rightarrow a=\frac{4}{3}b\Rightarrow \frac{3a}{b}=4

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