# If $u_n = (1+\frac{1}{n^2})\ldots(1+\frac{n}{n^2})$ then $\lim u_n=\sqrt{e}$?

I'm solving an exercise and I'm asked to prove that if

$$u_n = \left(1+\frac{1}{n^2}\right)\left(1+\frac{2}{n^2}\right)\left(1+\frac{3}{n^2}\right)\ldots\left(1+\frac{n}{n^2}\right)$$

then $\lim u_n = \sqrt{e}$.

This is after it asks me to show that $\forall x>0,$

$$x-\frac{1}{2}x^2<\ln(1+x)<x$$

which I did, but even with this result I'm not being able to even get how to use it to show $\lim u_n = \sqrt{e}$.

• Given the inequality, you may be tempted to take logarithm to obtain $$\sum_{k=1}^{n} \frac{k}{n^2} - \frac{k^2}{2n^4} < \log u_n < \sum_{k=1}^{n} \frac{k}{n^2}.$$ Can you proceed from here? – Sangchul Lee Mar 18 '18 at 20:51
• Yes! I was absolutely forgetting about the log properties to transform the sum into product. – Concept7 Mar 18 '18 at 21:01

$$\ln (u_n)=\sum_{k=1}^n\ln (1+\frac {k}{n^2} )$$ but for $k=1,2...n,$
$$\frac {k}{n^2}-\frac {k^2}{2n^4}<\ln (1+\frac {k}{n^2})<\frac {k}{n^2}$$
sum these inequalities and use the identities $$\sum_{k=1}^n\frac {k}{n^2}=\frac {n (n+1)}{2n^2}$$ and $$\sum_{k=1}^n\frac {k^2}{n^4}=\frac {n (n+1)(2n+1)}{6n^4}$$ Do not forget to squeeze.