Series Riddle

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

1. Anonymous11:58 PM

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

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

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) )

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.

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

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

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.)

1. Yeah. And I do so know pearantesthes.

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.

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.

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

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!

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.

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.