one doesn't have to find "an optimal N" (although finding such an N can be evidence of a certain...cleverness), just one that's "big enough".
in Zarathustra's example, the sequence {1/n} bounds {2^(-n)} from above, since {1/n} goes to 0 in the limit,
and both sequences are positive, we have a special case of the "squeeze theorem" at work: {0} < {2^(-n)} < {1/n}.
this is useful to remember when proving convergence: if the sequence is bounded by another sequence that makes the ε-algebra
easier, go ahead and find an N for the "bigger sequence".
of course the derivation of an optimal N is the case of 2^(-n) isn't that hard:
we want: |2^(-n)| = 2^(-n) < ε. so log2(2^(-n)) < log2(ε) (log base 2 is increasing so preserves order)
so -n < log2(ε), so we want n > -log2(ε) (the alternate form in emakarov's post no doubt comes from re-writing:
2^(-n) = e^(-nln(2)), which has the advantage of being in the form of the standard log function, which makes it more
transparent that ln is increasing, after all, it's derivative is 1/x, which is > 0 for all x in the domain of ln).