<s>[INST] 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 $$\text{lim}_{n\rightarrow\infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{1+1} = \frac{1}{2}\,._{\,\,\square}$$
 [/INST] 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.
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