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From: ecn-ec:ecn-pc:ecn-ed:vu@pur-ee.UUCP
Newsgroups: net.math
Subject: Re: An interesting identity
Message-ID: <152@ecn-ed.UUCP>
Date: Wed, 10-Aug-83 08:24:32 EDT
Article-I.D.: ecn-ed.152
Posted: Wed Aug 10 08:24:32 1983
Date-Received: Thu, 11-Aug-83 19:15:36 EDT
References: ihuxr.532
Lines: 40

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)