# How to calculate the limit of the sum f(1)+f(2)+....+f(n), if f(x)=ln[(x+2)/x], n->infinite?

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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:

f(1) + ....+f(n) = ln (n+1)(n+2)/2

We'll apply limit both sides:

lim

f(1) + ....+f(n) = lim ln (n+1)(n+2)/2

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