bhuber@sjuvax.UUCP (B. Huber) (11/27/85)
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? I suspect not, but cannot yet find a way to show that it is impossible. 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). Thus it would give a formula for the resistance through a circuit shaped like an n-cube (a problem which appeared in net.physics two months ago). 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)?
ghgonnet@watdaisy.UUCP (Gaston H Gonnet) (11/30/85)
> > For k>=0, let a(k) be 2^k / k (one over k, times the kth power of 2). ===================== you probably mean k>0 ===================== > Can you find the sum of a(k) as k ranges from 1 to n as a closed formula > in n? > A "closed formula" is a fuzzy concept, it depends on your set of primitive functions. E.g. Hn = sum(1/k,k=1..n) cannot be expressed in terms of + - * and /, but it can be expressed in terms of psi(x). I do not know any "closed form" for the above, if it is of any help, it has an asymptotic expansion like: sum( 2^k/k, k=1..n) = 2^(n+1) / n * ( 1 + 1/n + 3/n^2 + 13/n^2 + 75/n^4 + O(1/n^5) ) For a more general case ( sum( z^k/k, k=1..n ) ) there is also an asymptotic formula, see the "Handbook of Algorithms and Data Structures", Addison Wesley, p. 210 (II.1.7))
pumphrey@ttidcb.UUCP (Larry Pumphrey) (12/03/85)
> 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? > I suspect not, but cannot yet find a way to show that it is impossible. I suspect so, how about k=n 2^k (n+2) f(n) = sigma ----- = 2 - ----- k=0 k 2^n
pumphrey@ttidcb.UUCP (Larry Pumphrey) (12/03/85)
>> 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? >> I suspect not, but cannot yet find a way to show that it is impossible. > I suspect so, how about > k=n 2^k (n+2) > f(n) = sigma ----- = 2 - ----- > k=0 k 2^n I'll print my own correction, my erroneous "solution" was broadcast before I had a chance to retract it. Obviously, I interchanged the numerator and denominator which makes the problem trivial. Sorry about that :-(