Sequences and series (5)

We note that $S_n = \displaystyle \sum_{k=0}^\infty \frac n{(n+1)^k} = \frac n{1-\frac 1{n+1}} = \frac {n(n+1)}{n+1-1} = n+1$ . Then, we have:

$\begin{aligned} L & = \lim_{n \to \infty} \frac {S_1S_n+S_2S_{n-1}+S_3S_{n-2}+ \cdots + S_nS_1}{S_1^2+S_2^2+S_3^2+ \cdots + S_n^2} \\ & = \lim_{n \to \infty} \frac {\sum_{k=1}^n S_kS_{n-k+1}}{\sum_{k=1}^n S_k^2} \\ & = \lim_{n \to \infty} \frac {\sum_{k=1}^n (k+1)(n-k+2)}{\sum_{k=1}^n (k+1)^2} \\ & = \lim_{n \to \infty} \frac {\sum_{k=1}^n (k+1)(n-(k+1)+3)}{\sum_{k=1}^n (k+1)^2} \\ & = \lim_{n \to \infty} \frac {\sum_{k=1}^n (k+1)(n+3) - \sum_{k=1}^n (k+1)^2}{\sum_{k=1}^n (k+1)^2} \\ & = \lim_{n \to \infty} \frac {(n+3)\sum_{\color{#3D99F6}k=1} ^{\color{#3D99F6}n} ({\color{#3D99F6}k+1})}{\sum_{\color{#3D99F6}k=1}^{\color{#3D99F6}n} ({\color{#3D99F6}k+1})^2} - 1 \\ & = \lim_{n \to \infty} \frac {(n+3)\sum_{\color{#D61F06}k=2} ^{\color{#D61F06}n+1} \color{#D61F06}k}{\sum_{\color{#D61F06}k=2} ^{\color{#D61F06}n+1} {\color{#D61F06}k}^2} - 1 \\ & = \lim_{n \to \infty} \frac {\frac 12 (n+3)(n+1)(n+2)-1}{\frac 16 (n+1)(n+2)(2n+3)-1} - 1 \\ & = \frac {\frac 12}{\frac 26} - 1 = \frac 32 - 1 = \frac 12 = \boxed{0.5} \end{aligned}$