We'll have to substitute x by 1,2,...n, into the given expression of f(x).
f(1) = ln (1+2)/1 = ln 3/1
f(2) = ln (2+2)/2 = ln 4/2
f(3) = ln (3+2)/3 = ln 5/3
f(n-2) = ln (n-2+2)/(n-2) = ln n/(n-2)
f(n-1) = ln (n-1+2)/(n-1) = ln (n+1)/(n-1)
f(n) = ln (n+2)/n
We'll add f(1) + ....+f(n) = ln 3/1 + ln 4/2 + ln 5/3 + ... + ln n/(n-2) + ln (n+1)/(n-1) + ln (n+2)/n
Since the bases of logarithms are matching, we'll transform the sum into a product:
f(1) + ....+f(n) = ln (3/1)*(4/2)*(5/3)*...*[n/(n-2)]*[(n+1)/(n-1)]*[(n+2)/n]
We'll simplify and we'll get:
We'll apply limit both sides:
lim ln (n+1)(n+2)/2 = ln lim (n+1)(n+2)/2
Since the order of numeraor is bigger than the order of denominator, the limit is infinite.
lim [ f(1) + ....+f(n)] = infinite