proof of slower divergent series

Let us show that, if the series ${\sum }_{i=1}^{\mathrm{\infty }}{a}_i$ of positive terms is divergent, then Abel’s series

$$\sum _{i=1}^{\mathrm{\infty }}\frac{a_i}{{\sum }_{j=1}^i{a}_j}$$

also diverges.

Since the series ${\sum }_{i=1}^{\mathrm{\infty }}{a}_i$ diverges, we can find an increasing sequence ${(n_i)}_{i=0}^{\mathrm{\infty }}$ of integers such that

$$\sum _{j=1}^{n_{i+1}}{a}_j>2\sum _{j=1}^{n_i}{a}_j$$

for all $i$. By convention, set $n_0=0$. Then we can group the sum like so:

$$\sum _{i=1}^{n_m}\frac{a_i}{{\sum }_{j=1}^i{a}_j}=\sum _{i=0}^{m-1}\sum _{k=n_i+1}^{n_{i-1}}\frac{a_k}{{\sum }_{j=1}^k{a}_j}$$

Because ${\sum }_{j=1}^k{a}_j\le {\sum }_{j=1}^{n_{i-1}}{a}_j$, we have

$$\sum _{k=n_i+1}^{n_{i-1}}\frac{a_k}{{\sum }_{j=1}^k{a}_j}\ge \sum _{k=n_i+1}^{n_{i-1}}\frac{a_k}{{\sum }_{j=1}^{n_{i-1}}{a}_j}=\frac{{\sum }_{k=n_i+1}^{n_{i-1}}{a}_k}{{\sum }_{j=1}^{n_{i-1}}{a}_j}$$

By the way we chose the sequence ${(n_i)}_{i=0}^{\mathrm{\infty }}$, we have $2{\sum }_{k=n_i+1}^{n_{i-1}}{a}_k>{\sum }_{j=1}^{n_{i-1}}{a}_j$ and, hence,

$$\sum _{k=n_i+1}^{n_{i-1}}\frac{a_k}{{\sum }_{j=1}^k{a}_j}\ge \frac{{\sum }_{k=n_i+1}^{n_{i-1}}{a}_k}{{\sum }_{j=1}^{n_{i-1}}{a}_j}>\frac{1}{2}.$$

Therefore,

$$\sum _{i=1}^{n_m}\frac{a_i}{{\sum }_{j=1}^i{a}_j}>\sum _{i=1}^{n_m}\frac{1}{2}=\frac{m}{2},$$

so the sum diverges in the limit $m\to \mathrm{\infty }$.