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Prove 2^(1/4)*4^(1/8)*8^(1/16)*16^(1/32) ... ∞ = 2.
Alternative view : http://www.wolframalpha.com/input/?i=2%5E%281%2F4%29*4%5E%281%2F8%29*8%5E%281%2F16%29*16%5E%281%2F32%29+...+%E2%88%9E+%3D+2&dataset=&equal=Submit
 one year ago
 one year ago
row Group Title
Prove 2^(1/4)*4^(1/8)*8^(1/16)*16^(1/32) ... ∞ = 2. Alternative view : http://www.wolframalpha.com/input/?i=2%5E%281%2F4%29*4%5E%281%2F8%29*8%5E%281%2F16%29*16%5E%281%2F32%29+...+%E2%88%9E+%3D+2&dataset=&equal=Submit
 one year ago
 one year ago

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mukushla Group TitleBest ResponseYou've already chosen the best response.2
let's see what is its closed form \[\large \prod_{n=1}^{\infty} (2^n)^{\frac{1}{2^{n+1}}}=\prod_{n=1}^{\infty} (2)^{\frac{n}{2^{n+1}}}=2^{\sum_{n=1}^{\infty} \frac{n}{2^{n+1}}}=2\]so u just need to show that \[\large \sum_{n=1}^{\infty} \frac{n}{2^{n+1}}=1\]
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
make sense ???
 one year ago
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