<s>[INST] Calculate limit in use of integrals. Calculate limit in use of integrals:
$$ \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k+n}{3k^2+n^2+1} $$
Solution:
$$\sum_{k=1}^{n} \frac{k+n}{3k^2+n^2+1} = \frac{1}{n} \sum_{k=1}^{n} \frac{\frac{k}{n}+1}{3(k/n)^2+1 + n^{-2}} = \\
\frac{1}{n}\cdot \sum_{k=1}^{n} \frac{\frac{k}{n}+1}{3(k/n)^2+1 +n^{-2}} $$
I want use there approximation by Riemann integrals. But I don't know how to deal with $n^{-2}$
 [/INST] For any $\varepsilon>0$ and all $n>1/\sqrt\varepsilon$, we have
$$
\frac{k+n}{3k^2+n^2}\geq
\frac{k+n}{3k^2+n^2+1}\geq
\frac{k+n}{3k^2+(1+\varepsilon)n^2}.
$$
So we have
$$
\int_0^1\frac{x+1}{3x^2+1}\,\mathrm{d}x\geq\lim_{n\to\infty}\frac{k+n}{3k^2+n^2+1}\geq\int_0^1\frac{x+1}{3x^2+(1+\varepsilon)}\,\mathrm{d}x
$$
and now take $\varepsilon\to 0$.
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