# Given `f(x)=3/2x^{2/3}` and series `a_n=1/{1^{1/3}+1/2^{1/3}+cdots+1/n^{1/3}}` , show that `1/(k+1)^{1/3}<3/2(k+1)^{2/3}-3/2k^{2/3}<1/k^{1/3}` k >0

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You should notice that the function `f(x) = (3/2)root(3)(x^2)` is continuous, hence, the function is differentiable over an interval `(k,k+1).`

Since the function is differentiable over an interval `(k,k+1)` , hence, there is a value `c in (k,k+1)` such that:

`f(k+1) - f(k) = f'(c)(k+1 - k) => (f(k+1) - f(k))/(k+1 - k) = f'(c)`

You need to differentiate the function with respect to x such that:

`f'(x) = (3/2)*(2/3)*x^(2/3 - 1) => f'(x) = x^(-1/3) => f'(x) = 1/(x^(1/3))`

Hence, evaluating `f'(c)` yields:

`f'(c) = 1/(c^(1/3))`

You need to notice that `f'(c) = f(k+1) - f(k)` , hence, you need to evaluate `f(k+1) - f(k)` such that:

`f(k+1) - f(k) = (3/2)((k+1)^(2/3) - k^(2/3)) => f'(c) = (3/2)((k+1)^(2/3) - k^(2/3))` .

Since `f'(c) = 1/(c^(1/3)) => 1/(c^(1/3)) = (3/2)((k+1)^(2/3) - k^(2/3))` .

Notice that `c in (k , k+1) => k < c < k+1 => (k^(1/3)) < (c^(1/3)) < ((k+1)^(1/3)) => 1/(k^(1/3)) > 1/(c^(1/3)) > 1/((k+1)^(1/3))`

Substituting `(3/2)((k+1)^(2/3) - k^(2/3))` for `1/(c^(1/3))` yields:

`1/(k^(1/3)) >(3/2)((k+1)^(2/3) - k^(2/3)) > 1/ ((k+1)^(1/3))`

**Notice that using the properties of continuous and differentiable function `f(x) = (3/2)root(3)(x^2)` yields `1/(k^(1/3)) > (3/2)((k+1)^(2/3) - k^(2/3)) > 1/ ((k+1)^(1/3)),` for `k > 0` .**