已知各项都是正数的数列 {an}(n∈N∗)\left\{ {a}_{n}\right\} \left( {n \in {\mathbf{N}}^{ * }}\right){an}(n∈N∗) 的前 nnn 项和为 Sn{S}_{n}Sn ,且 Sn=an2+2an{S}_{n} = \frac{{a}_{n}}{2} + \frac{2}{{a}_{n}}Sn=2an+an2 ,则下列结论中正确的是 ( )A. {an}\left\{ {a}_{n}\right\}{an} 是单调递增数列B. Sn+Sn+2<2Sn+1{S}_{n} + {S}_{n + 2} < 2{S}_{n + 1}Sn+Sn+2<2Sn+1C. 1S1+1S2+⋯+1Sn>Sn+1−22\frac{1}{{S}_{1}} + \frac{1}{{S}_{2}} + \cdots + \frac{1}{{S}_{n}} > \frac{{S}_{n + 1} - 2}{2}S11+S21+⋯+Sn1>2Sn+1−2D. Sn−23Sn−2≤14lnn\frac{{S}_{n} - 2}{3{S}_{n} - 2} \leq \frac{1}{4}\ln n3Sn−2Sn−2≤41lnn