Series Riddle

1/1·2 + 1/2·4 + 1/3·8 + 1/4·16 + 1/5·32 + ... = ?

14 comments:

  1. 17.0, but with an extra fraction that can never quite be positioned. It's there, but rounding causes it to go into infinity.

    ReplyDelete
    Replies
    1. You and I must interpret my admittedly clumsy typographic idiom differently from oneanother; the result can hardly overtop 1.

      Delete
  2. = 2/1 + 4/2 + 8/3 + 16/4 + 32/5 ...
    = 2 + 2 + 2 2/3 + 4 + 6.4 + ...

    Since each one gets larger, I'm going with infinity.

    Or do you mean:
    = 1/2 + 1/8 + 1/24 + 1/64 + 1/160... ?
    Well, it pretty much stops growing after the first twenty or so terms, so 0.693147181...

    But I have no idea how to do it analytically.

    Sum [n=1:inf] (1 / (n * 2^n) )

    ReplyDelete
    Replies
    1. Indeed it is Log(2), as the derivative of this series expansion is that of 1/(1-x). Also he doesn't know parentheses.

      Delete
    2. Shiny I'm pretty sure Wabulon knows about parentheses. He's just lazy.

      Delete
    3. You and I had the same mismatch as Fetz and I.

      Delete
  3. ln(2).

    Andy: one way to prove it is to write down the Taylor expansion at infinity of (1/x)*(1/(x-1)) and integrate it to see that the sum from 1 to infinity of 1/(n*x^n) is the Taylor series at infinity for ln(x/(x-1)). Evaluating at x=2 gives the result.

    (Of course, it's slightly annoying/confusing that the Taylor series at infinity is really a Laurent series and not a power series at all.)

    ReplyDelete
    Replies
    1. Yeah. And I do so know pearantesthes.

      Delete
  4. shonk,

    Given the information that it is ln(2), I can probably do the Taylor series expansion and then reduce it to the form above. What I don't know how to do is the inverse. Did you guys just recognize 0.693... as ln(2) and go from there? The only irrationals I'd have so recognized would be sqrt(2), 1/sqrt(2), e, Pi, Pi/2.

    ReplyDelete
    Replies
    1. Yes, I definitely recognize 0.693... as ln(2) (for one thing, this is why the Rule of 70 works).

      As for working backwards, I immediately assumed that Wabulon's series was Laurent series with a variable in place of the 2. If you make that assumption then you immediately notice that the series gets nicer if you differentiate once, so that's what I did. What you get is practically the Taylor series at infinity for -1/(1-x) = 1/(x-1) (this is obvious if you've derived the Taylor series at 0 for 1/(1-x) as many times as I have); there's just an extra factor of 1/x. That gives that the derivative of Wabulon's series is the series for 1/(x(x-1)), so you just integrate to get the original function.

      Delete
    2. I did it your way; in fact, I chose the series to make that easy.

      Delete
  5. Where is Ludwig Wittgenstein when you need him?

    The first few items in a series don't tell us "how to go on." - there are uncountably many ways of doing so!

    ReplyDelete
    Replies
    1. But Kevin, surely Wittgenstein's point here Is that we do know how to go on, but that knowing is not itself a sort of rule or algorithm.

      Delete
    2. Absolutely true, a point I made to my father as a child (he loved series riddles). Yet, as in love and war, despite all philosophy and logic in our way, somehow we know how to proceed.

      Delete