# Given f(x)=1/x^2, prove that 1/(k+1)<f(k+1)-f(k)<1/k.

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We'll apply Lagrange's rule over the closed interval [k ; k+1], for the function f(x) = 1/x^2.

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

f(k+1) - f(k) = -2/x^3

c belongs to [k ; k+1] <=> k < c < k+1

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

We'll raise to square:

1/k^2 > 1/c^2 > 1/(k+1)^2

-2/k^3 < -2/c^3 < -2/(k+1)^3

But -2/c^3 = f(k+1) - f(k)

-2/k^3 <f(k+1) - f(k)<-2/(k+1)^3

**According to Lagrange's rule, the inequality 1/(k+1)<f(k+1)-f(k)<1/k is not true, instead, the inequality -2/k^3 <f(k+1) - f(k)<-2/(k+1)^3 is true, over the interval [k ; k+1].**