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English
Theorem
19
[219] Let \(πΌ=\limsup _{nββ}\sqrt[n]{|a_ n|}\) then
if \(πΌ{\lt}1\) the series \(β_{n=1}^β a_ n\) converges absolutely;
if \(πΌ=1\) nothing can be concluded;
if \(πΌ{\gt}1\) the series \(β_{n=1}^β a_ n\) does not converge, and also \(β_{n=1}^β |a_ n|\) diverges.
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