Derivative of Natural Logarithm Function/Proof 4/Lemma

Theorem
Let $\left\langle{f_n}\right\rangle_n$ be the sequence of real functions $f_n: \R_{>0} \to \R$ defined as:
 * $f_n \left({x}\right) = n \left({\sqrt [n] x - 1}\right)$

Let $k \in \N$.

Let $J = \left[{\frac 1 k \,.\,.\, k}\right]$.

Then the sequence of derivatives $\left\langle{ {f_n}'}\right\rangle_n$ converges uniformly to some real function $g: J \to \R$.

Proof
From Derivative of $n$th Root and Combination Theorem for Sequences:


 * $\forall n \in \N: \forall x \in J : D_x f_n \left({x}\right) = \dfrac {\sqrt [n] x} x$

From Closed Bounded Subset of Real Numbers is Compact, $J$ is compact.

Thus:

In particular, $\left\langle{ {f_n}'}\right\rangle_n$ is pointwise convergent to a continuous function on $J$.

For $x \in \left[{\frac 1 k \,.\,.\, 1 }\right]$:

So $\left\langle{ {f_n}' \left({x}\right)}\right\rangle_n$ is increasing when $x \in \left[{\frac 1 k \,.\,.\, 1}\right]$.

For $x \in \left[{1 \,.\,.\, k}\right]$:

So $\left\langle{ {f_n}' \left({x}\right)}\right\rangle_n$ is decreasing when $x \in \left[{\frac 1 k \,.\,.\, 1}\right]$.

Thus $\left\langle{ {f_n}' \left({x}\right)}\right\rangle_n$ is monotone for all $x \in \left[{\frac 1 k \,.\,.\, 1}\right] \cup \left[{1 \,.\,.\, k}\right] = J$.

From Dini's Theorem, $\left\langle{ {f_n}'}\right\rangle_n$ converges uniformly to $\dfrac 1 x$ on $J$.

Hence the result.