# 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.

• Do you mean $3n^4+n^2$, or $n^4+n^2$? Commented Mar 25, 2020 at 21:10
• So, what's wrong with $\sum \frac{1}{n}f(\frac{k}{n})$? Commented Mar 25, 2020 at 21:13
• It seems to be convergent.I use matlab for approximation and its value was growing fast. Commented Mar 25, 2020 at 21:38
• I fixed it, RogerI. It's $3n^4+n^2$ Commented Mar 26, 2020 at 0:42

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.
• I got it wrong in the header - there are $2n^4+n^2$ summonds (as i wrote in the body of the question), not $n^2$ Commented Mar 26, 2020 at 0:00
• @rotemaracky That's probably just a typo since your booklet confirms the result should be $5$
There's an error in the question since if we denote the sum by $$S$$ , then:
$$(5n^2+1)\cdot \frac{2n^4}{3n^4+n^2} and the $$LHS$$ diverges.