Simplifying ratio test with exponents $k+1$

Question

Question: Find the interval and radius of convergence. $$\sum_{k=1}^\infty\frac{(x-1)^k(k^k)}{(k+1)^k} .$$

I applied the ratio test.

$$r=\lim_{k\to\infty}\frac{(x-1)^{k+1}(k+1)^{k+1}}{(k+2)^{k+1}}\times\frac{(k+1)^k}{(x-1)^k(k^k)}\ .$$

I am stuck here. Please help me simplify.

Answer 1

it is easyier to apply Cauchy (n-th root) test.

$\sqrt[k]{\frac{|x-1|^k.k^k}{(k+1)^k}}=|x-1|\frac{k}{k+1} \to |x-1| <1$

Answer 2

$\bigg(\dfrac{k + 1}{k + 2}\bigg)^{k+1} = \dfrac1{\bigg(1 + \frac1{k+1}\bigg)^{k+1}} \to \dfrac1e,\quad$ and $\quad\bigg(\dfrac{k + 1}k\bigg)^k = \bigg(1 + \dfrac1k\bigg)^k \to e.\quad$ So you have $r = |x - 1| < 1 \iff 0 < x < 2$.