Evaluate $\lim_{n\to \infty}\frac{1}{\sqrt[4]{{n^4}+n+2}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}}$ I am new to analysis and I have no clue how to solve this limit. This is an exam problem from my analysis 1 course, there are one or two similar ones on the exam.

$$\lim_{n\to \infty}\frac{1}{\sqrt[4]{{n^4}+n+2}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}}$$

The only thing I tryed was this silly idea to rewrite it as one single fraction
and apply Stolz-Cesaro theorem, but it got way too messy so I doubt that is the way.

I can't find explanations generally on these limits of sequences of the type $\frac{1}{f(x_n)}+\cdots+\frac{1}{f(x_{n+k})}$ (I hope this is a good representation). Should series be involved in solving these kinds of limits ?

EDIT: The limit is supposed to be solved only with the knowledge prior to derivatives and integrals.

Thanks in advance
Hint  Assuming that the sum means
$$\lim_{n\to \infty}\frac{1}{\sqrt[4]{{n^4}+n+2}}+\frac{1}{\sqrt[4]{{n^4}+n+3}}+\frac{1}{\sqrt[4]{{n^4}+n+4}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}}$$

Use that 
$$\frac{1}{\sqrt[4]{{n^4}+5n-1}}+\frac{1}{\sqrt[4]{{n^4}+5n-1}}+\frac{1}{\sqrt[4]{{n^4}+5n-1}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}} \leq \\
\leq \frac{1}{\sqrt[4]{{n^4}+n+2}}+\frac{1}{\sqrt[4]{{n^4}+n+3}}+\frac{1}{\sqrt[4]{{n^4}+n+4}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+5n-1}} \leq \\
\leq \frac{1}{\sqrt[4]{{n^4}+n+2}}+\frac{1}{\sqrt[4]{{n^4}+n+2}}+\frac{1}{\sqrt[4]{{n^4}+n+2}}+\cdots+\frac{1}{\sqrt[4]{{n^4}+n+2}}$$
Also note that $\frac{1}{n+1}&lt;\frac{1}{n^4+n+k} &lt;\frac{1}{n}$
so $$ \sum_{k=1}^{4n-1} \frac{1}{n+1}&lt;S_n=\sum_{k=1}^{4n-1} \frac{1}{(n^4+n+k)^{1/4}} &lt;\sum_{k=1}^{4n-1} \frac{1}{n}.$$  So, we note that 
$$ \lim_{n \rightarrow \infty} \frac{4n-1}{n+1}=4= \lim_{n \rightarrow \infty} \frac{4n-1}{n}= \lim_{n \rightarrow \infty} S_n$$

When $n \rightarrow \infty$ the first term of the sequence $1/(n^4+n+2)^{1/4} \rightarrow 0$ it can well be neglected.

Note that $$\frac{1}{(n^4+n+k)^{1/4}} &gt; \frac{1}{n+1}$$ can be checked to be true for all $1&lt;k\le 4n-1$