I evaluate the following limit with Riemann integral
$$\lim _{n\to \infty }\frac{1}{n^2}\cdot \sqrt[n]{\left(n^2+1\right)\cdot \left(n^2+2\right)\cdot \left(n^2+5\right)\cdot \:\:\:...\:\:\:\cdot \left(2n^2-2n+2\right)}$$ That's equal with $$e^{\lim _{n\to \infty }\left(\frac{1}{n}\sum _{k=1}^n\:ln\left(1+\frac{\left(k^2-2k+2\right)}{n^2}\right)\right)}\:\:in\:this ,case\:when\:we\:calculate\:exponent,\:which\:will\:be\:the\:function?$$ I'm total confused $$how \:we\:see\:\frac{\left(k^2-2k+2\right)}{n^2}\:like\:x\:?\:for\:our\:function\:f\:\left(x\right)=ln(1\:+\:?)\:\:...\:and\:explain\:why\:we\:see\:like\:that$$ Please make me to understand that, I apreciate!
Put $\displaystyle u_n=\frac{1}{n}\sum_{k=1}^n \log(1+\frac{k^2-2k+2}{n^2})$, $\displaystyle v_n=\frac{1}{n}\sum_{k=1}^n \log(1+\frac{k^2}{n^2})$, $w_n=u_n-v_n$. Let $g(u)=\log(1+u)$, $u\in [0,1]$. As $g$ is continuous on the compact $I=[0,1]$, it is uniformly continuous on $I$. Hence if $\varepsilon>0$ is given, there exists $\alpha>0$ such that for any $u,v\in I$ verifying $|u-v|<\alpha$, we have $|g(u)-g(v)|<\varepsilon$. Take $\displaystyle u=\frac{k^2-2k+2}{n^2}$ and $\displaystyle v=\frac{k^2}{n^2}$, we have $|u-v|\leq 2/n <\alpha$ for $n$ large. Show that this imply that $w_n\to 0$, and now it is easy to finish.