Calculate limit in use of integrals

Question

Calculate limit in use of integrals

$$ \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1+n}{3k^2+n^2} $$

My attempt:

$$\sum_{k=1}^{n} \frac{1+n}{3k^2+n^2} = \frac{1}{n} \sum_{k=1}^{n} \frac{\frac{1}{n}+1}{3(k/n)^2+1} = \\ \frac{1}{n}\cdot (1/n + 1) \sum_{k=1}^{n} \frac{1}{3(k/n)^2+1} $$ Ok, I know that when I am taking limit I should replace (from aproximation theorem) $$ \sum_{k=1}^{n} \frac{1}{3(k/n)^2+1}$$ with $$ \int_{0}^{1} \frac{1}{1+3x^2}$$ but I still don't know what have I do (and why) with $$ \frac{1}{n}\cdot (1/n + 1) $$ part. In many solutions we just ignore part $\frac{1}{n}$ but I don't know why and there where I have little more 'difficult' expression like $ \frac{1}{n}\cdot (1/n + 1) $ I completely don't know what should I do... $$ $$

Answer 1

$$\frac{1}{n}\left(1+\frac{1}{n}\right)\sum...=\underbrace{\frac{1}{n}\sum...}_{\to \int...}+\underbrace{\frac{1}{n}\underbrace{\left(\frac{1}{n}\sum_{}...\right)}_{\to \int...}}_{\to 0}\to \int...$$

Answer 2

Just ignore the $1$ in the numerator as it is obviously less than $\infty$.

And put $\frac{1}{n}=dx$ & $\frac{k}{n}=x$

So your integration becomes $$\int_0^1\frac{dx}{3x^2+1}=\frac{tan^{-1}\sqrt{3}}{\sqrt{3}}$$

Answer to your question, what to do with $\frac{1}{n}(\frac{1}{n}+1)$ as $(\frac{1}{n}+1)\to1$ as $n\to\infty$.

Remember, when factors are in multiplication, partial limit can be applied.

Hope this will be helpful!