Convergence of $\frac{1}{n}\sum_{i=1}^{n}\left[\frac{\left(\log\left(1+i/n\right)\right)^2}{1+i/n}\right]$ Sequence of real numbers
$$S_n=\frac{1}{n}\sum_{i=1}^{n}\left[\frac{\left(\log\left(1+\frac{i}{n}\right)\right)^2}{1+\frac{i}{n}}\right]$$
Does $\lim\limits_{n \to \infty} S_n$ exist? If so, compute the value.
My Solution:
A)
According to Cauchy's first theorem on limits.
This limit will be same as $$\lim_{n \to \infty}\left[\frac{(\log(1+\frac{n}{n}))^2}{1+\frac{n}{n}}\right] =\frac{(\log(2))^2}{2}$$
B) $S_n$ can also be written as $$\int_{0}^{1}\frac{(\log(1+x))^2}{1+x}dx=\frac{(\log(2))^3}{3}$$
Two answers by two methods. What am I missing here? Is this sequence not convergent?

the B) methods is true. 
$$\lim_{n\to\infty}\dfrac{1}{n}\sum_{i=1}^{n}\left(\dfrac{(\ln{1+\dfrac{i}{n}})^2}{1+\dfrac{i}{n}}\right)=\lim_{n\to\infty}\dfrac{1}{n}\sum_{i=1}^{n}f\left(\dfrac{i}{n}\right)=\int_{0}^{1}f(x)dx$$
where $$f(x)=\dfrac{\left(\ln{(1+x)}\right)^2}{1+x}$$
