$\frac{n^{-h} - 1}{h} = -\log n + O(|h|(\log n)^2)$ for $|h|\log n \leq 1$

Question

I'm trying to prove the continuity of $\zeta(s)$. As part of this proof, I've arrived at a term $$ \frac{n^{-h} - 1}{h} $$ which I want to bound. I wanted to see if it was possible to show that this term is $-\log n + O(|h|(\log n )^2)$ if $|h|\log n \leq 1$, but I'm having trouble introducing a $\log n $ quantity on the left hand side to do this.

Answer 1

The bound can be proven by taking a Taylor Expansion. The term is already a polynomial in terms of $n$, so the expansion is taken in terms of $h$. Rewrite the fraction as $$\left(\frac{1}{h}\right)(n^{-h} - 1)$$ and take the Taylor Expansion of $n^{-h}$. The constant $-1$ will cancel out with the first term of the expansion, the divisor $h$ will cancel out with one power of $(-h)$ in all later terms, and the bound will be given by the remainder.