# proof of slower divergent series

Let us show that, if the series $\sum_{i=1}^{\infty}a_{i}$ of positive terms is divergent, then Abel’s series

 $\sum_{i=1}^{\infty}\frac{a_{i}}{\sum_{j=1}^{i}a_{j}}$

also diverges.

Since the series $\sum_{i=1}^{\infty}a_{i}$ diverges, we can find an increasing sequence $(n_{i})_{i=0}^{\infty}$ 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 $\sum_{j=1}^{k}a_{j}\leq\sum_{j=1}^{n_{i-1}}a_{j}$, we have

 $\sum_{k=n_{i}+1}^{n_{i-1}}\frac{a_{k}}{\sum_{j=1}^{k}a_{j}}\geq\sum_{k=n_{i}+1% }^{n_{i-1}}\frac{a_{k}}{\sum_{j=1}^{n_{i-1}}a_{j}}=\frac{\sum_{k=n_{i}+1}^{n_{% i-1}}a_{k}}{\sum_{j=1}^{n_{i-1}}a_{j}}$

By the way we chose the sequence $(n_{i})_{i=0}^{\infty}$, we have $2\sum_{k=n_{i}+1}^{n_{i-1}}a_{k}>\sum_{j=1}^{n_{i-1}}a_{j}$ and, hence,

 $\sum_{k=n_{i}+1}^{n_{i-1}}\frac{a_{k}}{\sum_{j=1}^{k}a_{j}}\geq\frac{\sum_{k=n% _{i}+1}^{n_{i-1}}a_{k}}{\sum_{j=1}^{n_{i-1}}a_{j}}>\frac{1}{2}.$

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 $m\to\infty$.

Title proof of slower divergent series ProofOfSlowerDivergentSeries 2013-03-22 15:08:42 2013-03-22 15:08:42 rspuzio (6075) rspuzio (6075) 9 rspuzio (6075) Proof msc 40A05