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SolomonZelman
 one year ago
I have a question about an understanding of a concept. It is easy  not hard.
SolomonZelman
 one year ago
I have a question about an understanding of a concept. It is easy  not hard.

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0Say, 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
 one year ago
Best ResponseYou've already chosen the best response.0the \(\large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }~\sqrt[n]{A_n}}\) is the geometric ratio  simply speaking. It is expolring how the series behaves as we get to so to speak "infinitiith" terms.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0For example \(\large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }~\sqrt[n]{\frac{\rm x^n}{\rm (5n+3)^n}}}\) \(\large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }~\frac{\rm x}{\rm 5n+3}=r}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0and 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
 one year ago
Best ResponseYou've already chosen the best response.0tnx for any views, and out:)
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