SolomonZelman
  • SolomonZelman
I have a question about an understanding of a concept. It is easy - not hard.
Mathematics
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chestercat
  • chestercat
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SolomonZelman
  • SolomonZelman
Say, I got any series \(\large\color{black}{ \displaystyle \sum_{{\rm n}=k}^{\infty}{\rm A}_{\rm n} }\) where the root test for converges in practical, for instance (in red) \(\large\color{black}{ \displaystyle \sum_{{\rm n}=k}^{\infty}\frac{x^{\rm n}}{(5{\rm n }+3)^{\rm n}} }\) (or if you can give a better example, please do) why does the nth root test show that series converges if the (absolute value of the) result is less than 1?
SolomonZelman
  • SolomonZelman
the \(\large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }~\sqrt[n]{A_n}}\) is the geometric ratio - simply speaking. It is expolring how the series behaves as we get to so to speak "infinitiith" terms.
SolomonZelman
  • SolomonZelman
For example \(\large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }~\sqrt[n]{\frac{\rm x^n}{\rm (5n+3)^n}}}\) \(\large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }~\frac{\rm x}{\rm 5n+3}=r}\)

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SolomonZelman
  • SolomonZelman
and if that expression \(\large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }~\frac{\rm x}{\rm 5n+3}}\) has an absolute value (forgot absolute value of the limit) then it converges, just as any elementary geom. series would with |r|<1
SolomonZelman
  • SolomonZelman
tnx for any views, and out:)

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