limit of sequence

(a) Consider a(n+1) - a(n) = 1/(n+1) - ln(n+2) + ln(n+1)
= 1/(n+1) - ln[(n+2)/(n+1)]
= 1/(n+1) - ln[1/(n+1) + 1]
>= 1/(n+1) - 1/(n+1) = 0
Therefore a(n) is an increasing sequence
Consider b(n+1) - b(n) = 1/(n+1) - ln(n+1) + ln(n)
= 1/(n+1) + ln[n/(n+1)]
= 1/(n+1) + ln[-1/(n+1) + 1]
<= 1/(n+1) + [-1/(n+1)] = 0
Therefore b(n) is an decreasing sequence
(b) Since b(n) is a decreasing sequence, b(1) > b(n) or 1 > b(n)
Also a(n) is an increasing sequence, a(1) < a(n) or 1 - ln(2) < a(n)
b(n) - a(n) = ln(n+1) - ln(n)
b(n) - a(n) = ln(1 + 1/n) > 0
b(n) > a(n)
So 1 > b(n) > a(n) > 1 - ln(2)
So a(n) is bounded upwards and b(n) is bounded downwards
Therefore both a(n) and b(n) are convergent
lim n-> ∞ [b(n) - a(n)] = lim n-> ∞ ln(1 + 1/n) = ln(1) = 0
lim n-> ∞ b(n) - lim n-> ∞ a(n) = 0
lim n-> ∞ b(n) = lim n-> ∞ a(n)