# slower divergent series

###### Theorem.

If

 $\displaystyle a_{1}\!+\!a_{2}\!+\!a_{3}\!+\cdots$ (1)

is a diverging series with positive , then one can always form another diverging series

 $s_{1}\!+\!s_{2}\!+\!s_{3}\!+\cdots$

with positive such that

 $\displaystyle\lim_{n\to\infty}\frac{s_{n}}{a_{n}}=0.$ (2)

Proof.   Let  $S_{n}=a_{1}\!+\!a_{2}\!+\cdots+\!a_{n}$  be the $n^{\mathrm{th}}$ partial sum of (1).  Then we have

 $a_{n}=S_{n}\!-\!S_{n-1}=(\sqrt{S_{n}}\!+\!\sqrt{S_{n-1}})(\sqrt{S_{n}}\!-\!% \sqrt{S_{n-1}}).$

We set  $s_{1}:=\sqrt{S_{1}}$  and

 $s_{n}:=\frac{a_{n}}{\sqrt{S_{n}}\!+\!\sqrt{S_{n-1}}}=\sqrt{S_{n}}\!-\!\sqrt{S_% {n-1}}$

for  $n=2,\,3,\,4,\,\ldots$  Then the of the series

 $\sum_{n=1}^{\infty}s_{n}=\sqrt{S_{1}}\!+\!\sum_{n=1}^{\infty}(\sqrt{S_{n+1}}\!% -\!\sqrt{S_{n}})$

apparently are positive.  This series is however divergent, because the sum of its $n$ first is equal to $\sqrt{S_{n}}$ which grows without bound along with $n$ since (1) diverges.  For this reason we also get the result (2).

Remark.Niels Henrik Abel has presented a simpler example on such series $s_{1}\!+\!s_{2}\!+\!s_{3}\!+\cdots$:

 $1\!+\!\frac{a_{2}}{a_{1}\!+\!a_{2}}\!+\!\frac{a_{3}}{a_{1}\!+\!a_{2}\!+\!a_{3}% }\!+\!\frac{a_{4}}{a_{1}\!+\!a_{2}\!+\!a_{3}\!+\!a_{4}}\!+\cdots$
Title slower divergent series SlowerDivergentSeries 2013-03-22 15:08:27 2013-03-22 15:08:27 pahio (2872) pahio (2872) 13 pahio (2872) Theorem msc 40A05 SlowerConvergentSeries NonExistenceOfUniversalSeriesConvergenceCriterion