Now, ln(n+1) > ln(n) as ln is a strictly increasing function, therefore:
0 < a(n) < 1 for all n and as such a(n) is both upper and lower bounded.
B. In the interval [n , n+1] 1/x is, of course, strictly decreasing, and as such its integral is greater than its right sum for any partition of [n , n+1]. Therefore we can consider the 'dumb' partition P1 = {n , n+1} and we'll have:
Integral [n , n+1] 1/x dx > 1 /(n+1) * (n+1 - n) = 1 / (n+1) or
ln(n+1) - ln(n) > 1 / (n+1)
and therefore 1 / (n+1) + ln(n) - ln(n+1) < 0
Now, a(n+1) - a(n) =
(1 + 1/2 + ... + 1/n + 1/(n+1) - ln(n+1)) - (1 + 1/2 + ... + 1/n - ln(n)) =
= 1 / (n+1) + ln(n) - ln(n+1), which is negative (see right above)
Since, therefore, a(n+1) - a(n) < 0 , a(n) is a strictly decreasing sequence, and since we already proved it is lower bounded, it converges to some real number y (greek gamma), which is called Euler's constant.
Note that it's still undetermined (open problem) if Euler's constant is a rational or irrational number.