Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

  • This Question is Open
  1. henpen
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    .

  2. mukushla
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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 \[\large \sum_{n=1}^{\infty} \frac{n}{2^{n+1}}=1\]

  3. mukushla
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    make sense ???

  4. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy