Let $a_n>0$ for $n \geq 1$ and let series: $\sum_{n=1}^{\infty}a_n$ diverge. Let $S_n=a_1+a_2+...+a_n > 1$ for $n \geq 1$ Prove that the series: $$\sum_{n=1}^{\infty}\frac{a_{n+1}}{S_n \ln S_n}$$ diverges and the series : $$\sum_{n=1}^{\infty}\frac{a_{n}}{S_n \ln^2 S_n}$$ converges. (Using the famous criteria I pressume). For some reason I cannot see which criteria to use here that would be effective. On first glance Dalambere's criteria seems applicable but am not sure at all how to use it.

First 
$$
\int_{S_{n}}^{S_{n+1}}\dfrac{dx}{x\ln(x)}\leqslant\int_{S_{n}}^{S_{n+1}}\dfrac{dx}{S_{n}\ln(S_{n})}=\dfrac{1}{S_{n}\ln(S_{n})}\int_{S_{n}}^{S_{n+1}}dx=\dfrac{a_{n+1}}{S_{n}\ln(S_{n})}
$$
for $S_{n+1}\geqslant S_n$ and $\dfrac1{x\ln(x)}\downarrow$. So 
$$
\sum\limits_{n=1}^{N}\dfrac{a_{n+1}}{S_{n}\ln(S_{n})}\geqslant\sum\limits_{n=1}^{N}\int_{S_{n}}^{S_{n+1}}\dfrac{dx}{x\ln(x)}=\int_{S_{1}}^{S_{N+1}}\dfrac{dx}{x\ln(x)}=\ln(\ln(S_{N+1}))-\ln(\ln(S_{1}))\to\infty 
$$
for $S_N\to\infty$. So $\sum\limits_{n=1}^{\infty}\dfrac{a_{n+1}}{S_{n}\ln(S_{n})}$ diverges.
Second
$$
\int_{S_{n-1}}^{S_{n}}\dfrac{dx}{x\ln^2(x)}\geqslant\int_{S_{n-1}}^{S_{n}}\dfrac{dx}{S_{n}\ln^2(S_{n})}=\dfrac{1}{S_{n}\ln^2(S_{n})}\int_{S_{n-1}}^{S_{n}}dx=\dfrac{a_{n}}{S_{n}\ln^2(S_{n})}
$$
for $S_{n}\geqslant S_{n-1}$ and $\dfrac1{x\ln^2(x)}\downarrow$. So 
$$
\sum\limits_{n=1}^{N}\dfrac{a_{n}}{S_{n}\ln^2(S_{n})}\leqslant\sum\limits_{n=1}^{N}\int_{S_{n-1}}^{S_{n}}\dfrac{dx}{x\ln^2(x)}=\int_{S_{1}}^{S_{N}}\dfrac{dx}{x\ln^2(x)}=\dfrac{1}{\ln(S_{1})}-\dfrac{1}{\ln(S_{N+1})}\to\dfrac{1}{\ln(S_{1})} 
$$
for $S_N\to\infty$. So $\sum\limits_{n=1}^{\infty}\dfrac{a_{n}}{S_{n}\ln^2(S_{n})}$ converges.
