let me know if I am correct? - to check convergence and find limit If not please correct me, also another method to solve is very welcome.
For given series $$S_n = \sum_{i=1}^{n} \frac{(\log(n+i)-\log{n})^2}{n+1}$$
To test its convergence, $$S_n = \sum_{i=1}^{n} \frac{(\log(n+i)-\log{n})^2}{n+1} =\sum_{i=1}^{n} \frac{(\log(1+\frac{i}{n}))^2}{n+1}$$
$$ \implies S_n \leq \sum_{i=1}^{n} \frac{(\frac{i}{n})^2}{n+1}$$
$$ \implies S_n \leq \frac{\frac{n(2n+1)(n+1)}{6}}{n^2(n+1)}$$
$$ \implies \lim_{n\to\infty} S_n \leq \lim_{n\to\infty}\frac{\frac{n(2n+1)(n+1)}{6}}{n^2(n+1)}$$
$$ \implies \lim_{n\to\infty} S_n \leq \frac{1}{3}$$
Thus limit converges
To find the limit of sum, I changes sum into reimann sum
$$S_n =\sum_{i=1}^{n} \frac{(\log(1+\frac{i}{n}))^2}{n+1}$$
$$ \implies S_n =\int_{1}^{2} \lim_{n\to\infty} \frac{n}{n+1}(\log(1+x))^2$$
$$ \implies S_n =\int_{1}^{2} (\log(1+x))^2$$
Using Integration by parts
$$ \implies S_n =2(\log2)^2-2\log2+1 $$
You used the wrong integral.
$S_n \to \int_0^1 \log^2(1 + x) dx = 2 (\log(2) - 1)^2≈0.18832 $.
You took $1+x$ into account twice instead of once.
This now agrees with your upper bound.