**phr@ucbvax.ARPA (Paul Rubin)** (05/29/85)

Andrew Nash, Peter Royappa, and a few other people answered the last head scratcher, which was to demonstrate a bijection between (0,1) and [0,1]. The solutions were all pretty hard for me to follow, but the idea was the same in each: [0,1] is (0,1) with two extra points, 0 and 1. So you can map everything to itself, except you have to first get rid of these points, by mapping them to (say) 1/2 and 1/3 respectively. Then you must get rid of 1/2 and 1/3, by sending them to (say) 1/4 and 1/5, and so on down the line. One such mapping f from [0,1] to (0,1) then is f(x) = x, unless x = 0 or x = 1/n for some integer n; f(x) = 1/(n+2) otherwise. ====================================================================== Now the new one: Give an example of an infinite sequence a_0, a_1, ... so that the series infinity ____ \ a_n does not converge, but /___ n = 0 infinity ____ \ a_n / (1 + n a_n) converges. /___ n = 0

**egs@epsilon.UUCP (Ed Sheppard)** (05/29/85)

How about _ | | 1 if i=2^j-1 for some j in N a[i] = | | 0 otherwise |_ so that the sequence a looks like this 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, ... The sum of a[i] diverges, but the sum of a[i]/(1+i*a[i]) is 2. Ed Sheppard Bellcore