How to conjugate $\sqrt[m]{n+b}-\sqrt[m]{n}$, so we get $\frac{b}{\text{stuff}}$

Question

$b\in\mathbb{R}^{+}$

$n\in\mathbb{N}_0$

$m\in\mathbb{N}_2$

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

Answer 1

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, $$ (n+b)^{\large\frac{\normalsize1}m}-n^{\large\frac{\normalsize1}m}=\frac{b}{\sum\limits_{k=1}^mn^{\large\frac{k-1}m}(n+b)^{\large\frac{m-k}m}}\tag{3} $$

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Simple proof of the formula for the sum of a geometric sequence via a telescoping series $$ \begin{align} (1-r)\sum_{k=1}^nr^{k-1} &=\sum_{k=1}^n\left(r^{k-1}-r^k\right)\\ &=\sum_{k=0}^{n-1}r^k-\sum_{k=1}^nr^k\\[9pt] &=1-r^n\tag{4} \end{align} $$

Answer 2

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.

Answer 3

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.