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

Also note that $\frac{1}{n+1}<\frac{1}{n^4+n+k} <\frac{1}{n}$
so $$ \sum_{k=1}^{4n-1} \frac{1}{n+1}<S_n=\sum_{k=1}^{4n-1} \frac{1}{(n^4+n+k)^{1/4}} <\sum_{k=1}^{4n-1} \frac{1}{n}.$$  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 $$\frac{1}{(n^4+n+k)^{1/4}} > \frac{1}{n+1}$$ can be checked to be true for all $1<k\le 4n-1$ 
