I'm trying to prove the following inequality: $$\left\lvert\frac{x^h -1}{h}-\log(x)\right\rvert\leq \frac{|h||\log(x)|^2 }{2}\exp(|h\log(x)|), \quad 0<|h|<1.$$
What I've tried: By the Taylor series, we have $x^h=1+h\log(x)+\frac{h^2 \log^2(x)}{2}+\cdots $. Then we have $$\frac{x^h -1}{h}-\log(x)=\frac{h^2 \log^2(x)}{2}+\cdots .$$ Thus, we have $$\left\lvert\frac{x^h -1}{h}-\log(x)\right\rvert=\left\lvert\frac{h^2 \log^2(x)}{2}+\cdots \right\rvert\leq \frac{|h||\log(x)|^2 }{2}+\cdots .$$ But the results are not the same. I was wondering if someone could tell me how to prove the first inequality.
You forgot to divide by h:$$\left\rvert\frac{x^h-1}{h}-\log{(x)}\right\rvert=\left\rvert\frac{h\log^2{(x)}}{2}+...\right\rvert$$Now notice that we can write the RHS like $$\left\rvert\frac{1}{h}\sum\limits_{k=2}^{\infty}\frac{(h\log{(x)})^k}{k!}\right\rvert\le\frac{1}{h}\sum\limits_{k=2}^{\infty}\left\rvert\frac{(h\log{(x)})^k}{k!}\right\rvert=\frac{1}{h}\frac{|h||\log^2{(x)}|}{2}*2\sum\limits_{k=2}^{\infty}\frac{|h|^{k-1}|\log^{k-2}{(x)}|}{k!}\\=\frac{|h||\log^2{(x)}|}{2}*2\sum\limits_{k=2}^{\infty}\frac{(|h||\log{(x)}|)^{k-2}}{k!}\le\frac{|h||\log^2{(x)}|}{2}\sum_{k=0}^{\infty}\frac{(|h||\log{(x)}|)^{k}}{k!}\\=\frac{|h||\log^2{(x)}|}{2}\exp{(|h\log{(x)}}|)$$