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JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Sequences and Series

If

\frac{1}{1 ^{4}} + \frac{1}{2 ^{4}} + \frac{1}{3 ^{4}} + \dots \infty = \frac{π ^{4}}{90}

,

\frac{1}{1 ^{4}} + \frac{1}{3 ^{4}} + \frac{1}{5 ^{4}} + \dots = α

,

\frac{1}{2 ^{4}} + \frac{1}{4 ^{4}} + \frac{1}{6 ^{4}} + \dots \infty = β

,
then

\frac{α}{β}

is equal to

(A)23
(B)18
(C)15✓ Correct
(D)14

Answer: (C)

Solution

The even-power sum is

β = \sum_{n \geq 1} \frac{1}{( 2 n ) ^{4}} = \frac{1}{16} \sum_{n \geq 1} \frac{1}{n ^{4}} = \frac{1}{16} \cdot \frac{π ^{4}}{90}

. The odd-power sum is the whole minus the even part,

α = \frac{π ^{4}}{90} - \frac{1}{16} \cdot \frac{π ^{4}}{90} = \frac{15}{16} \cdot \frac{π ^{4}}{90}

. Dividing,

\frac{α}{β} = \frac{15/16}{1/16} = 15

.

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option

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