${\displaystyle h_{n}=\varphi (n){\frac {v_{n}}{v_{n+1}}}-\varphi (n+1)={\frac {\varphi (n)u_{n}}{u_{n+1}(1-\alpha _{n+1}u_{n+1})}}-\varphi (n+1),}$

${\displaystyle h_{n}=\varphi (n){\frac {u_{n}}{u_{n+1}}}-\varphi (n+1)+\alpha _{n+1}{\frac {\varphi (n)u_{n}}{1-\alpha _{n+1}u_{n+1}}}.}$

${\displaystyle \varphi (n)={\frac {u_{1}+u_{2}+\cdots +u_{n}}{u_{n}}}={\frac {S_{n}}{u_{n}}};}$

${\displaystyle \lim \varphi (n)u_{n}=\infty ,\qquad \varphi (n){\frac {u_{n}}{u_{n+1}}}-\varphi (n+1)=-1;}$

${\displaystyle \lim h_{n}=\pm \infty }$

${\displaystyle u_{n}={\frac {1}{n}},\quad ={\frac {1}{n\log n}},\quad ={\frac {1}{n\log n\log ^{2}n}},\ldots ,\alpha _{n}=k=\mathrm {cost.}$ ;}

${\displaystyle {\begin{aligned}&\sum {\frac {1}{n}}\left(1-{\frac {k}{1}}\right)\left(1-{\frac {k}{2}}\right)\cdots \left(1-{\frac {k}{n}}\right),\\\\&\sum {\frac {1}{n\log n}}\left(1-{\frac {k}{2\log 2}}\right)\left(1-{\frac {k}{3\log 3}}\right)\cdots \left(1-{\frac {k}{n\log n}}\right),\\\\&\sum {\frac {1}{n\log n\log ^{2}n}}\left(1-{\frac {k}{2\log 2\log ^{2}2}}\right)\left(1-{\frac {k}{3\log 3\log ^{2}3}}\right)\cdots \left(1-{\frac {k}{n\log n\log ^{2}n}}\right),\\&.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;.\;\end{aligned}}}$