Compute: $\lim_{n\to\infty}\sum_{i=1}^{2n^4+n^2}\frac{5n^2+1}{n^4+i}$

Question

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.

Answer 1

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.

Answer 2

There's an error in the question since if we denote the sum by $S$ , then:

$(5n^2+1)\cdot \frac{2n^4}{3n^4+n^2}<S$ and the $LHS$ diverges.