<s>[INST] Compute: $\lim_{n\to\infty}\sum_{i=1}^{2n^4+n^2}\frac{5n^2+1}{n^4+i}$ The question is to find $\lim_{n\to\infty} \left ( \frac{5n^2+1}{n^4+1}+\frac{5n^2+1}{n^4+2}+\frac{5n^2+1}{n^4+3}+...\frac{5n^2+1}{3n^4+n^2} \right )$
The answer is 5, according to the booklet.
I tried to bring the sum to the form $ \sum \frac{1}{n}f(\frac{k}{n})$ in order to use Rieman sum. Also I tried to apply Cauchy's first theorem on limits. Didn't get me closer to something helpful.
 [/INST] Let $a_n=\frac{5n^2+1}{n^4+1}+\frac{5n^2+1}{n^4+2}+\frac{5n^2+1}{n^4+3}+...+\frac{5n^2+1}{n^4+n^2}$, then notice
$$
a_n \geq n^2\cdot \frac{5n^2+1}{n^4+n^2}=\frac{5+\frac{1}{n^2}}{1+\frac{1}{n^2}}
$$
and similarly
$$
a_n \leq n^2 \cdot  \frac{5n^2+1}{n^4+1} = \frac{5+\frac{1}{n^2}}{1+\frac{1}{n^4}}.
$$
Now just use the squeeze theorem.
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