# Verify if 1/(k+1)<f(k+1)-f(k)<1/k, if the function f isĀ given by f(x)=ln x, k>0?

### 1 Answer | Add Yours

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).**