lew@ihuxr.UUCP (08/04/83)
I stumbled across the following identity: sum i=1 to n of (-1)^(i+1) * C(n,i) * 1/i equals sum i=1 to n of 1/i Can anybody provide a reference for this, or show its equivalence to some basic identity? Surely Euler knew this! When I tried to prove it inductively by taking S(n+1) - S(n), I got sum i=1 to n of (-1)^(i+1) * C(n,i) * 1/(n+1-i) equals 0, n even ; 2/(n+1), n odd ... so this doesn't really help. Lew Mammel, Jr. ihuxr!lew
ecn-ec:ecn-pc:ecn-ed:vu@pur-ee.UUCP (08/10/83)
Induction *IS* the way. Then we arrive at: sum i=1 to n (-1)^(i+1) *C(n,i) * 1/(n+1-i) equals 0, n even; 2/(n+1) n odd. as equivalent to the given identity. Now, remark that C(n,i)/(n+1-i) = C(n+1, i) / (n+1) substitute in: sum i=1 to n (-1)^(i+1) * C(n+1,i) / (n+1) equals 0 n even 2/(n+1) n odd. Add (-1)^(n+2) / (n+1) both sides: sum i=1 to (n+1) (-1)^(i+1) * C(n+1,i) / (n+1) equals (-1)^n / (n+1) n even ( 2 + (-1)^n ) / (n+1),n odd equals 1 / (n+1) for all n . Multiply both sides by (n+1), we have that the given identity is equivalent to: sum i=1 to (n+1) (-1)^(i+1) * C(n+1,i) = 1 Put n' = n+1 and drop the primes: sum i=1 to n (-1)^(i+1) * C(n,i) = 1 This series looks so simple that it has got to be in some elem stat text. I have not taken any stat course at all, so that I don't know where it can be found. Hao-Nhien Vu (pur-ee!vu or better: pur-ee!norris)