Evaluate $\lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}$ 
$$\lim_{n \to
 \infty}\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}$$

How to consider it?

Here's my solution. Please correct me if I'm wrong.
First, let's introduce and prove a result, which will be applied soon. That is

$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n^2-n+nk}}=2.$$

Consider using the squeeze theorem. Notice that
\begin{align*} \frac{1}{\sqrt{n+1-k}+\sqrt{n-k}}\geq\frac{1}{2\sqrt{n+1-k}}&\geq\frac{1}{\sqrt{n+1-k}+\sqrt{n+2-k}}, \end{align*}
which implies
\begin{align*} \sqrt{n+1-k}-\sqrt{n-k}\geq\frac{1}{2\sqrt{n+1-k}}&\geq\sqrt{n+2-k}-\sqrt{n+1-k}. \end{align*}
Therefore
$$2\geq\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+n-nk}}\geq\frac{2\left(\sqrt{n+1}-1\right)}{\sqrt{n}}\to 2(n \to \infty)$$
As per the squeeze theorem, the conclusion is followed.
Now, let's manage to solve the present problem. Since $\sqrt[k]{k}$ is decreasing for $k\geq 2$, and converges to $1$ as $k \to \infty$. Thus
$$\forall \varepsilon>0,\exists N>0,\forall n>N,\sqrt[k]{k}\leq1+\varepsilon.$$
Hence
\begin{align*}
\sum_{k=1}^{n}\frac{1}{\sqrt{n^2-n+nk}}&\leq\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}\\&=\sum_{k=1}^{N}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}+\sum_{k=N+1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}\\
&\leq\sum_{k=1}^{N}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}+(1+\varepsilon)\sum_{k=N+1}^{n}\frac{1}{\sqrt{n^2-n+nk}}\\
&\leq\sum_{k=1}^{N}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}+(1+\varepsilon)\sum_{k=1}^{n}\frac{1}{\sqrt{n^2-n+nk}}.
\end{align*}
Take the limits as $n \to \infty$ of both sides. We obtain
$$2\leq \lim_{n \to \infty}\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}\leq 2(1+\varepsilon).$$
Since $\varepsilon$ is arbitary, we can conclude 
$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}=2,$$
which is what we want.
