Legitimacy of a solution 
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*Problem


$$a_n=(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})$$
Find $\lim_{n\rightarrow\infty}a_n$


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*My solution


We have $$\ln(a_n)=\ln(1+\frac{1}{n^2})+\ln(1+\frac{2}{n^2})+...\ln(1+\frac{n}{n^2})=$$ $$=\frac{1}{n^2}+o(\frac{1}{n^2})+\frac{2}{n^2}+o(\frac{2}{n^2})+...+\frac{n}{n^2}+o(\frac{n}{n^2})$$ $$=\frac{n(n+1)}{2n^2}+no(\frac{1}{n})$$
Taking the limit when $n\rightarrow\infty$ we get
$$\lim_{n\rightarrow\infty}ln(a_n)=\frac{1}{2}$$ or $\lim_{n\rightarrow\infty}a_n=\sqrt{e}$


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*My question: I'm pretty sure that my result is right, but I don't know if my proof is rigorous, especially in the part I take:
$$o(\frac{1}{n^2})+o(\frac{2}{n^2})+...+o(\frac{n}{n^2})=no(\frac{1}{n})$$


It seems true, but is it rigorous?
 A: What you need is : $$\sum_{k=1}^n o(\frac{k}{n^2} )= o(1)$$
The proof is simple, for $n$ large enough (depending on $\epsilon>0$) we have $\left|o\left(\frac{k}{n^2} \right)\right|<\frac{\epsilon }{n}$, sum to get : $$\left|\sum_{k=1}^n o(\frac{k}{n^2} )\right|\leq \sum_{k=1}^n \left|o(\frac{k}{n^2} )\right| \leq \epsilon$$
This finishes the proof. For an alternate solution you could use :$$ \forall x> 0 \ : \ x>\ln(1+x)>x-\frac{x^2}{2}.$$
This way you get rid of all $o$ and $\epsilon$ headaches.
A: In the same track as your, let us consider $\log(1+\frac{i}{n^2})$ when $n$ is large. Using one more term, the Taylor expansion is $$\log(1+\frac{i}{n^2})=\frac{i}{n^2}-\frac{i^2}{2 n^4}+O\left(\left(\frac{1}{n}\right)^5\right)$$ Now, sum. For the first portion, we get $$\frac{n+1}{2 n}$$ for the second $$-\frac{(n+1) (2 n+1)}{12 n^3}$$ in which we see that the degree of numerator is $2$ and degree of denominator is $3$. Adding more terms will make the difference of degrees increasing by one unit.  So, the limit of the sum is what you wrote. If we add the two sums, we then find for the sum of the logarithms, $$\frac{(n+1) \left(6 n^2-2 n-1\right)}{12 n^3}$$ the development of which being $$\frac{1}{2}+\frac{1}{3 n}+O\left(\left(\frac{1}{n}\right)^2\right)$$
