Verify if 1/(k+1)

Since the function f(x) is continuously and it could be
differentiated, we'll apply Lagrange's rule, over a closed interval [k ;
k+1].

According to Lagrange's rule, we'll
have:

f(k+1) - f(k) = f'(c)(k+1 -
k)

c belongs to the interval [k ;
k+1].

f'(x) = 1/x => f'(c) =
1/c

We'll eliminate like terms inside brackets and we'll
have:

f(k+1) - f(k) = f'(c), where f'(c) =
1/c

f(k+1) - f(k) = 1/c

Since
c belongs to [k ; k+1], we'll write:

k < c <
k+1

1/k > 1/c >
1/(k+1)

But f(k+1) - f(k) =
1/c

1/k > f(k+1) - f(k) > 1/(k+1)
q.e.d.

According to Lagrange's rule 1/k
> f(k+1) - f(k) > 1/(k+1).