To infinity and below

Algebra Level 4

∑ k = 2 ∞ [ exp ⁡ 2 ( − 2 + ( ∑ i = 1 ∞ 1 k i ) − 1 ) ] − 1 = ? \large \sum_{k=2}^\infty \left [\exp_2 \left(-2 + \left(\sum_{i=1}^\infty \dfrac1{k^i} \right)^{-1} \right) \right]^{-1} = \, ?

Notation : exp ⁡ A ( B ) \exp_A (B) denotes the exponential function A B A^B .


The answer is 4.

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M K by M K , 22, USA

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2 solutions

Chew-Seong Cheong
Jun 29, 2016

S = ∑ k = 2 ∞ 1 2 − 2 + 1 ∑ i = 1 ∞ 1 k i = ∑ k = 2 ∞ 1 2 − 2 + 1 1 k − 1 = ∑ k = 2 ∞ 1 2 − 2 + k − 1 = ∑ k = 2 ∞ 1 2 k − 3 = 2 ∑ k = 0 ∞ 1 2 k = 2 ( 1 1 − 1 2 ) = 4 \begin{aligned} S & = \Large \sum_{k=2}^\infty \frac 1{2^{-2+\frac 1{\sum_{i=1}^\infty \frac 1{k^i}}}} = \sum_{k=2}^\infty \frac 1{2^{-2+\frac 1{\frac 1{k-1}}}} \\ & = \sum_{k=2}^\infty \frac 1{2^{-2+k-1}} = \sum_{k=2}^\infty \frac 1{2^{k-3}} = 2 \sum_{k=0}^\infty \frac 1{2^k} = 2 \left( \frac 1{1-\frac 12} \right) = \boxed{4} \end{aligned}

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M K
Jun 29, 2016

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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