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 $$\text{lim}_{n\rightarrow\infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{1+1} = \frac{1}{2}\,._{\,\,\square}$$
 A: 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}
A: Hint: Use the fact that $a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+...+ab^{k-2}+b^{k-1})$ where $a=\sqrt[k]{n+1}$ and $b=\sqrt[k]{n}$
or 
$$0<a-b=\frac{a^k-b^k}{(a^{k-1}+a^{k-2}b+...+ab^{k-2}+b^{k-1})}<\frac{1}{kb^{k-1}}=\frac{1}{k\sqrt[k]{n^{k-1}}}$$
A: Hint: It might be useful applying mean-value theorem for continuous function $f(x)=\sqrt[k]{x}$ on $[1,\infty)$ then there exists $\xi\in[n,n+1]$ such that
$$\sqrt[k]{n+1}-\sqrt[k]{n}=\dfrac{1}{k\sqrt[k]{\xi}}\leq\dfrac{1}{k\sqrt[k]{n}}$$
A: Write $$b_n=\frac{\sqrt[k]{1+\frac{1}{n}}-\sqrt[k]{1}}{\frac{1}{\sqrt[k]{n}}}=\frac{\sqrt[k]{1+\frac{1}{n}}-\sqrt[k]{1}}{\frac{1}{{n}}}\frac{1}{n^{1-1/k}}$$
The first part is a difference quotient, hence $$\lim_{n\to \infty} b_n= f'(1)\lim_{n\to \infty}n^{1/k-1}$$
where $f(x)=\sqrt[k]{x}$.  We can use calculus to compute $f'(x)=\frac{1}{k}x^{(\frac{1}{k}-1)}$ and hence $f'(1)=\frac{1}{k}$, a finite constant.  Hence the limit will be zero if $\lim_{n\to \infty}n^{1/k-1}=0$, which is zero for $k>1$.  This allows us to also calculate that $b_nn^\alpha$ converges, provided $\alpha+\frac{1}{k}-1\le 0$.
Note that this works even for $k$ not an integer.
