Prove $\frac{\sqrt[n]k+1}{\sqrt[n]k-1}=\frac{2}{k-1}\left(\frac{k+1}{2}+\sum_{j=1}^{n-1}\sqrt[n]{k^j}\right)$

Question

Ramanujan's nested radical

(1)

$$\frac{\sqrt[4]5+1}{\sqrt[4]5-1}=\frac{1}{2}\left(3+\sqrt[4]5+\sqrt[4]{5^2}+\sqrt[4]{5^3}\right)$$

We proposed the general formula of Ramanujan's equation above

(2)

$$\frac{\sqrt[n]k+1}{\sqrt[n]k-1}=\frac{2}{k-1}\left(\frac{k+1}{2}+\sum_{j=1}^{n-1}\sqrt[n]{k^j}\right)$$

Setting $k=5$ and $n=4$ yeild (1)

Prove (2)

Where $(k,n)\ge2$

Answer 1

$$\begin{align}\frac{x+1}{x-1}&=\frac{(x+1)(x^{n-1}+\cdots+1)}{(x-1)(x^{n-1}+\cdots+1)}\\ &=\frac{x^n+2(x^{n-1}+\cdots+x)+1}{x^n-1} \end{align}$$

Letting $x=\sqrt[n]k$ and you get:

$$\frac{\sqrt[n]k+1}{\sqrt[n]k-1}=\frac{k+2\left(\sum_{j=1}^{n-1}\sqrt[n]{k^j}\right)+1}{k-1}$$

Answer 2

Hint: For $a \ne 1$ you have the (finite) geometric sum formula $$ \sum_{j=1}^{n-1} a^j = \frac{a^n - 1}{a - 1} - 1 \, . $$ With $a = \sqrt[n]{k}$ your result follows easily.