How to prove that $1+x-(1+x^{1+a})^{\frac1{1+a}}=1-\frac{x^{-a}}{1+a}+O(x^{-(1+2a)})$ as $x→+∞$ for $a>0$

Question

$$1+x-(1+x^{1+a})^{\frac1{1+a}}=1-\frac{x^{-a}}{1+a}+O(x^{-(1+2a)}),\quad x → +∞,\ a>0.$$

It was provided to me kindly by a contributor here - and is much better than the one I had arrived at empirically. But how do I go about proving it - a reference would be great, a simple explanation better!

Even for $x>5$ the approximation is really useful to me in the evaluation of probabilities based on the integral of the LHS, for which I could not find an easy to manage form for use in various integrals.

All being well, a hope to use the approximation in a published work, so I will reference any answer: guidance on referencing would also be useful!

Jamie

Answer 1

Using the approximation $(1+y)^{\alpha} = 1 +\alpha y + O(y^2)$ as $y\to 0$, $$ \begin{align*} (1+x^{1+a})^{\frac{1}{1+a}} &= x\left(1 + x^{-a-1}\right)^{\frac{1}{1+a}} \\ &=x\left(1 + \frac{x^{-1-a}}{1+a} + O\left(x^{-2-2a}\right)\right) \\ & = x + \frac{x^{-a}}{1+a}+ O\left(x^{-1-2a}\right) \end{align*} $$

Therefore,

$$ 1+x - (1+x^{1+a})^{\frac{1}{1+a}} = 1 - \frac{x^{-a}}{1+a}+ O\left(x^{-1-2a}\right) $$

as $x\rightarrow \infty$.

Answer 2

$$y=\left(x^{a+1}+1\right)^{\frac{1}{a+1}}\implies \log(y)={\frac{1}{a+1}}\log\left(x^{a+1}+1\right)$$ $$\log\left(x^{a+1}+1\right)=(a+1)\log(x)+\log(1+\epsilon)\qquad\text{with} \qquad \epsilon=x^{-(a+1)}$$

$$\log(y)=\log(x)+\frac{1}{a+1}\Big[\epsilon -\frac{\epsilon ^2}{2}+O\left(\epsilon ^3\right) \Big]$$

$$y=x \Big[1+\frac{1}{a+1}\epsilon -\frac{a }{2 (a+1)^2}\epsilon ^2+O\left(\epsilon ^3\right)\Big]$$

$$1+x-y=1-\frac{x \epsilon }{a+1}+\frac{a x \epsilon ^2}{2 (a+1)^2}+O\left(\epsilon ^3\right)$$ Make now $\epsilon=x^{-(a+1)}$ to get your approximation and even better if you wish.