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From: bhuber@sjuvax.UUCP (B. Huber)
Newsgroups: net.math
Subject: Re: summation in closed form (2^k/k)
Message-ID: <2637@sjuvax.UUCP>
Date: Thu, 12-Dec-85 16:08:09 EST
Article-I.D.: sjuvax.2637
Posted: Thu Dec 12 16:08:09 1985
Date-Received: Sat, 14-Dec-85 08:14:24 EST
References: <7512@watdaisy.UUCP> <9600037@uiucdcsp>
Organization: St. Joseph's University, Phila. PA.
Lines: 51

>  I think you should have checked your formula out before you post it.
> 
>   The formula you've posted is for Sum{ k / 2^k }  not Sum { 2^k / k}.
> 
>   ...
 
> -Ben Leimkuhler

You are quite correct to ask me to check out my formula, or any formula for
that matter, before posting it.  As has been kindly pointed out by another
respondent, I carelessly did not restrict k to be nonzero in the expression.
I am happy that that was an unimportant mistake, easily corrected.
	Your reference to a 'formula', though, mystifies me.  My copy of the
original posting says only:

>For k>=0, let a(k) be  2^k / k  (one over k, times the kth power of 2).
>Can you find the sum of a(k) as k ranges from 1 to n as a closed formula
>in n?  
...
>The answer is of some interest.  It would yield a closed formula for the
>sums of reciprocals of entries in Pascal's triangle (summed over individual
>rows).   ...

>A nice little mathematical puzzle is to find the exact relationship between
>the two sums which I have mentioned.  In particular, can you find it using
>only elementary techniques (algebraic, not transcendental; formal power
>series, etc., not allowed)?

I can find no formula that needs  'checking'.  But I'll give it here, both
to show how beautiful it is, and to ask others with an algebraic bent whether
you can prove it true, using only algebraic techniques.
 __________________________________________________________________________
|The sum of a(k), as k ranges from one to n+1,  all divided by a(n+1),     |
|equals      								   |
|the sum of the reciprocals of the binomial coefficients C(n,k) as k ranges|
|from 0 to n.								   |
|__________________________________________________________________________|

As an illustration, here are the first several sums:

n=1: (2/1) / (2/1)                            =       1/1
n=2: (2/1  +  4/2) / (4/2)                    =     1/1  +  1/1
n=3: (2/1  +  4/2  +  8/3) / (8/3)            =   1/1  +  1/2  +  1/1
n=4: (2/1  +  4/2  +  8/3  +  16/4) / (16/4)  = 1/1  +  1/3  +  1/3  +  1/1
...

My derivation of this is rather messy, and transcendental in flavor (solving
boundary value problems, e.g.).  Hence the desire for a nicer proof.

By the way Ben, I really would like to see what this has to do with the sum
of {k / 2^k}.