Show $b_n=\sqrt[k]{n+1}-\sqrt[k]{n}$ converges towards $0$ for all $k \geq 2$. I'm looking for help with (b) and (c) specifically. I'm posting (a) for completeness.
(a) Show convergence for $a_n=\sqrt{n+1}-\sqrt{n}$ towards $0$ and test $\sqrt{n}a_n$ for convergence.
(b) Show $b_n=\sqrt[k]{n+1}-\sqrt[k]{n}$ converges towards $0$ for all $k \geq 2$.
(c) For which $\alpha\in\mathbb{Q}_+$ does $n^\alpha b_n$ converge?

I'm pretty sure I solved (a). I have proven the convergence of $a_n$ by using the fact that $$\sqrt{n}<\sqrt{n+1}\leq\sqrt{n}+\frac{1}{2\sqrt{n}}$$ which holds true since $$(\sqrt{n}+\frac{1}{2\sqrt{n}})^2=n+1+\frac{1}{4n}\geq n+1\,.$$
This gives us $$0<\sqrt{n+1}-\sqrt{n}\leq\frac{1}{2\sqrt{n}}$$
and after applying the squeeze theorem with noting that $\frac{1}{2\sqrt{n}}\longrightarrow0$ we can tell that also $a_n\longrightarrow0$.
Now $x_n=\sqrt{n}a_n=\sqrt{n}(\sqrt{n+1}-\sqrt{n})$.
We have \begin{align*}\sqrt{n}(\sqrt{n+1}-\sqrt{n})&=\sqrt{n}\sqrt{n+1}-\sqrt{n}\sqrt{n}\\&=\sqrt{n(n+1)}-n\\&=\sqrt{n^2+n}-n\\&=\frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n}\\&=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}\\&=\frac{n}{\sqrt{n^2+n}+n}\\&=\frac{n}{n\sqrt{1+\frac{1}{n}}+n}\\&=\frac{1}{\sqrt{1+\frac{1}{n}}+1}\end{align*}
and hence since the harmonic sequence $\frac{1}{n}$ converges towards 0 we have $$\text{lim}_{n\rightarrow\infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{1+1} = \frac{1}{2}\,._{\,\,\square}$$

In order to solve (b), let $a=\sqrt[k]{n+1}$ and $b=\sqrt[k]n$. Then\begin{align}\sqrt[k]{n+1}-\sqrt[k]n&=a-b\\&=\frac{(a-b)(a^{k-1}+a^{k-2}b+\cdots+ab^{k-2}+b^{k-1})}{a^{k-1}+a^{k-2}b+\cdots+ab^{k-2}+b^{k-1}}\\&=\frac{a^k-b^k}{a^{k-1}+a^{k-2}b+\cdots+ab^{k-2}+b^{k-1}}\\&=\frac1{\sqrt[k]{n+1}^{k-1}+\sqrt[k]{n+1}^{k-2}\sqrt[k]n+\cdots+\sqrt[k]{n+1}\sqrt[k]n^{k-2}+\sqrt[k]n^{k-1}}\end{align}and therefore\begin{align}\lim_{n\to\infty}\sqrt[k]{n+1}-\sqrt[k]n&=\lim_{n\to\infty}\frac1{\sqrt[k]{n+1}^{k-1}+\sqrt[k]{n+1}^{k-2}\sqrt[k]n+\cdots+\sqrt[k]{n+1}\sqrt[k]n^{k-2}+\sqrt[k]n^{k-1}}\\&=0.\end{align}
