Use Taylor's theorem to derive an expression for $\log\left(1-\frac{1}{(n+1)^\alpha}\right)-\log\left(1-\frac{1}{n^\alpha}\right)$

Question

In Some asymptotic expressions in the theory of numbers, the author asserts on p.119 that

$\log\left(1-\frac{1}{(n+1)^\alpha}\right)-\log\left(1-\frac{1}{n^\alpha}\right)=\frac{\alpha}{n^{1+\alpha}-n}-\frac{1}{2}*\frac{\alpha}{(\xi^{\alpha+1}-\xi)^2}\bigg((\alpha+1)\xi^{\alpha}-1\bigg)$

where $n$ is an integer, $0<\alpha<1$, and $\xi$ is some real number between $n$ and $n+1$. How is this expression derived? The author says this is due to Taylor's theorem, so I presume the expression on the right is either a difference of two Taylor series or a difference of two Taylor polynomials, but I am still confused as to how they're obtaining this result. Some manipulation of the logarithms on the left hand side produces

$$\log\bigg(\frac{((n+1)^{\alpha}-1) \, n^{\alpha}}{(n+1)^{\alpha} \, (n^{\alpha}-1)}\bigg)$$

but it's unclear to me whether a Taylor expansion of this produces the result we're looking for.

Answer 1

Taylor's theorem refers to the fact that the difference between a function $f(x)$ and its $k$th degree Taylor polynomial centered at $a$ tends to zero as $x \to a$. There are explicit forms for the remainder. The Lagrange form of the remainder says that there exists $\xi$ between $a$ and $x$ such that (using $T_{k,a}(x)$ for the $k$th degree Taylor polynomial centered at $a$) $$ f(x) - T_{k,a}(x) = \frac{f^{(k+1)}(\xi)(x-a)^{k+1}}{(k+1)!} $$ The author applies this to $f(x) = \log\left(1 - \frac{1}{x^\alpha}\right)$, $a=n$, $x=n+1$, and $k=1$.

To see why, first write $$ f(x) = \log\left(\frac{x^\alpha -1}{x^\alpha}\right) = \log(x^\alpha -1) - \alpha \log x $$ So that $$ f'(x) = \frac{\alpha x^{\alpha - 1}}{x^\alpha -1} - \frac{\alpha}{x} = \frac{\alpha x^{\alpha} - \alpha(x^\alpha - 1)}{x^{\alpha + 1}-a} = \frac{\alpha}{x^{\alpha +1} - x} $$

Furthermore, $$ f''(x) = - \frac{\alpha}{\left(x^{\alpha+1}-x\right)^2}((\alpha+1)x^\alpha -1) $$ By Taylor's theorem, there exists $\xi$ such that $n \leq \xi \leq n+1$ and \begin{align*} f(n+1) &= f(n) + f'(n)(1) + \frac{1}{2} f''(\xi)(1)^2 \\\implies \log\left(1 - \frac{1}{(n+1)^\alpha}\right) &= \log\left(1 - \frac{1}{n^\alpha}\right) + \frac{\alpha}{n^{\alpha +1} - n} - \frac{1}{2} \frac{\alpha}{\left(\xi^{\alpha+1}-\xi\right)^2}((\alpha+1)\xi^\alpha -1) \end{align*} as claimed.