Sum of a series - 1

Algebra Level 2

∑ k = 1 ∞ 1 ( 2 k ) 3 = 1 n \sum_{k=1}^{\infty} \dfrac{1}{(2^k)^3 } = \frac 1n

The equation above holds true for integer n n . Find n n .


The answer is 7.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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2 solutions

∑ k = 1 ∞ 1 ( 2 k ) 3 = ∑ k = 1 ∞ 1 8 k = 1 8 ∑ k = 0 ∞ 1 8 k = 1 8 ⋅ 1 1 − 1 8 = 1 7 \sum_{k=1}^\infty \frac 1{(2^k)^3} = \sum_{k=1}^\infty \frac 1{8^k} = \frac 18 \sum_{k=0}^\infty \frac 1{8^k} = \frac 18 \cdot \frac 1{1-\frac 18} = \frac 1{\boxed 7}

4 Helpful 2 Interesting 3 Brilliant 0 Confused
Chris Lewis
Oct 5, 2020

∑ k = 1 ∞ 1 ( 2 k ) 3 = ∑ k = 1 ∞ 1 8 k = ( ∑ k = 0 ∞ 1 8 k ) − 1 = 1 1 − 1 8 − 1 = 1 7 \sum_{k=1}^{\infty} \frac{1}{\left(2^k \right)^3}=\sum_{k=1}^{\infty} \frac{1}{8^k}=\left( \sum_{k=0}^{\infty} \frac{1}{8^k} \right)-1=\frac{1}{1-\frac18}-1=\frac17

2 Helpful 1 Interesting 1 Brilliant 0 Confused

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