Another head scratcher, and solution to the last one.

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Paul Rubin

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May 29, 1985, 1:14:45 AM5/29/85
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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

Ed Sheppard

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May 29, 1985, 1:44:18 PM5/29/85
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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

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