Quandaries and Queries Name: Dane Who is asking: Other Level: All Question: Something I noticed fooling around with a calculator about 30 years ago. Considering e = 2.718281828459045.... Using Window's Calculator you will find 1.111 = 2.8531167... 1.01101 = 2.731861... 1.0011001 = 2.71964085... 1.000110001 = 2.71841774... 1.00001100001 = 2.7182954... 1.00000110000011 = 2.178231875... 1.000000110000001 = 2.178219643... There apears to be a pattern. My conjecture is: 1.'infinite number of zeros'11'infinite number of zeros'1 = e. I have no idea if this is true and no idea on how to go about proving my conjecture one way or another. If you extend the pattern using a calculator you will find the answer eventually starts diverging from e. I think this is due to calculations being lost in the floating point round off fuzz inherent in all computers. Any insight you may have would be enlightment. Thank You Dane Hi Dane, That's a really nice observation. One of the facts about the number e is that the sequence (1+1)1, (1+ 1/2)2, (1+ 1/3)3, (1+ 1/4)4,... approaches e. That is the limit, as n approaches infinity, of (1 + 1/n)n is e Some authors take this fact as the definition of e and some authors define e some other way and then prove this fact as a theorem. Your sequence is (almost) every tenth term of the sequence (1 + 1/n)n. Your sequence is (1 + 1/n)n+1 where n is 10, 100, 1000, ... Since the sequence (1 + 1/n)n approaches e, the subsequence of every tenth term approaches e also. You multiply the terms of this sequence by the sequence (1+ 1/n). But the limit of the sequence (1+ 1/n) is 1 and hence the limit of your sequence is 1e = e. Penny Go to Math Central