Proving the convergence of a sequence

[Baron]

Prove that the sequence {1+(1/2)^n} converges to 1

The answer key said to pick N = max (1, ceiling of log_2 (1/ε) ) for any ε > 0

so that if n > N, then | 1+(1/2)^n - 1 | = (1/2)^n < ε

I understand why N was picked to be the ceiling of log_2 (1/ε) but why is N chosen to be the maximum of that and 1. Is it because N must be a natural number and if ε < 0.1, the ceiling of log_2 (1/ε) is negative?

If that is the case, is 1 arbitrary? Could 1 have been 0.8 or 2? Is there any specific reason the answer key chose 1?

 

[HallsofIvy]

Frankly, I don't see any point in saying "the larger of 1 and \(\displaystyle log_2(1/\epsilon)\)" either. Since N must be a positive integer, of course, it doesn't need to be said that it is "larger than 1"! It is sufficient to say "take N to be an integer larger than \(\displaystyle log_2(1/\epsilon)\)".

 

[Baron]

Frankly, I don't see any point in saying "the larger of 1 and \(\displaystyle log_2(1/\epsilon)\)" either. Since N must be a positive integer, of course, it doesn't need to be said that it is "larger than 1"! It is sufficient to say "take N to be an integer larger than \(\displaystyle log_2(1/\epsilon)\)".

If ε was picked to be 0.1, then N = -1. Could that be why we need to say that N has to be larger than 1 so that for any ε > 0, N would be a natural number.