Limit of $\sum \dfrac{1}{1^2+2^2+\cdots+n^2}$

Assuming the formula $$u_n=24(H_n-H_{2n+1})+\frac{6}{n+1} + 18$$ is correct, one has $$H_n=\ln n+\gamma+O(1/n)$$ and so $$u_n=24\ln\frac{n}{2n+1}+O(1/n)+\frac{6}{n+1} + 18$$ which means that $$\lim_{n\to\infty}u_n=-24\ln2+18.$$

Since you did prove that $$u_n=24(H_n-H_{2n+1})+\dfrac{6}{n+1} + 18$$ we could even go beyond the limit itself using the asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12p^2}+O\left(\frac{1}{p^4}\right)$$ Use it twice and continue with Taylor series for large values of $n$ to get $$u_n=18-24 \log (2)-\frac{3}{2n^2}+\frac{3}{n^3}+O\left(\frac{1}{n^4}\right)$$which shows the limit and also how it is approached.

I also allows a quick evaluation of $u_n$ even for small values on $n$. For example $$u_{10}=\frac{13114793}{9699690}\approx 1.35208$$ while the above truncated series would give $\frac{4497}{250}-24 \log (2)\approx 1.35247$.

You already did most of the work actually. Using your $u$, we can plug in approximations for $H_n$. Using the approximation of $$H_n = \ln(n) + \gamma + O\left( \frac{1}{n} \right)$$

we can get that $$u_n \approx 24(\ln(n) + \gamma - \ln(2n+1) - \gamma) +\frac{6}{n+1}+18 = 24\ln\left( \frac{n}{2n+1} \right) + \frac{6}{n+1}+18$$

The $6/(n+1)$ term approaches $0$. Since $\frac{n}{2n+1}$ approaches $1/2$, the limit is $$24\ln(1/2) + 18 = 18-24\ln(2) \approx 1.364$$