How to conjugate $\sqrt[m]{n+b}-\sqrt[m]{n}$, so we get $\frac{b}{\text{stuff}}$ $b\in\mathbb{R}^{+}$
$n\in\mathbb{N}_0$
$m\in\mathbb{N}_2$
I want to multiply
$$\sqrt[m]{n+b}-\sqrt[m]{n}$$ in such a way that we get
$$\frac{b}{\text{stuff}}$$
(since n+b-n=b)
Therefore I need to find a suiting conjugate to get rid of the roots in the nominator.
I want to learn the method behind it, so that I can do it by hand with an explicit $m$.
Could you point me to a good explanation on how to get this factor?
I am not interested in the bare solution, I want to understand it.
Thanks
 A: The formula for the sum of a geometric sequence is
$$
\sum_{k=1}^mr^{k-1}=\frac{1-r^m}{1-r}\tag{1}
$$
Set $\displaystyle r=\left(\frac{n}{n+b}\right)^{\large\frac1m}$ then multiply $(1)$ by $(n+b)^{\large\frac{m-1}m}$ to get
$$
\sum_{k=1}^mn^{\large\frac{k-1}m}(n+b)^{\large\frac{m-k}m}=\frac{(n+b)-n}{(n+b)^{\large\frac{\normalsize1}m}-n^{\large\frac{\normalsize1}m}}\tag{2}
$$
Therefore,
$$
(n+b)^{\large\frac{\normalsize1}m}-n^{\large\frac{\normalsize1}m}=\frac{b}{\sum\limits_{k=1}^mn^{\large\frac{k-1}m}(n+b)^{\large\frac{m-k}m}}\tag{3}
$$

Simple proof of the formula for the sum of a geometric sequence via a telescoping series
$$
\begin{align}
(1-r)\sum_{k=1}^nr^{k-1}
&=\sum_{k=1}^n\left(r^{k-1}-r^k\right)\\
&=\sum_{k=0}^{n-1}r^k-\sum_{k=1}^nr^k\\[9pt]
&=1-r^n\tag{4}
\end{align}
$$
A: We can write $\sqrt[m]{n+b} $ as $ \sqrt[m]{n} \times \sqrt[m]{1+\frac{b}{n}} $. Now we can expand $\sqrt[m]{1+\frac{b}{n}} $ using standard series expansion as $ 1+ \frac{bm}{n} +\frac{m(m-1)b^{2}}{2!n^{2}}...$ and now our expression becomes $\sqrt[m]{n} ( 1+ \frac{bm}{n} +\frac{m(m-1)b^{2}}{2!n^{2}}...)$. This when subtracted with $\sqrt[m]{n}$ gives us $\sqrt[m]{n} ( \frac{bm}{n} +\frac{m(m-1)b^{2}}{2!n^{2}}...)$. If we divide by $\sqrt[m]{n} $ we thus get a function $ f(b) $. Hope it helps.
A: Let me show you how to do it for $m=3$, and you can do the general case.
I’m going to use the identity $A^3-B^3=(A-B)(A^2+AB+B^2)$, substituting $(n+b)^{1/3}$ for $A$ and $n^{1/3}$ for $B$, to get
$$
b=(n+b)-n=\bigl((n+b)^{1/3}-n^{1/3}\bigr)\bigl[(n+b)^{2/3}+(n+b)^{1/3}n^{1/3}+n^{2/3}\bigr]\,,
$$
and you can finish it off. Note that the formula you get is not very neat.
