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graydarl
Group Title
Can someone hlep me with this math problem? serie : an=(alpha*n)/(n+1), n>=1 (alpha is from Rreal number). I must determine if the series is monotone and bouded thank you very much
 one year ago
 one year ago
graydarl Group Title
Can someone hlep me with this math problem? serie : an=(alpha*n)/(n+1), n>=1 (alpha is from Rreal number). I must determine if the series is monotone and bouded thank you very much
 one year ago
 one year ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i would try with \(\alpha =1\) first
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
in this case you would get \(a_n=\frac{n}{n+1}\) which is bounded below by \(1\) since \(\frac{n}{n+1}<1\) for all \(n\) and \(\lim_{n\to \infty}\frac{n}{n+1}=1\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i meant "bounded above by one" not below!
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
it is monotone increasing as you can check by taking the derivative and seeing that is always positive
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
then try the general case with \(\alpha\)
 one year ago

Fazeelayaz Group TitleBest ResponseYou've already chosen the best response.0
yes alpha* n/(n+1) limi\[\lim_{x \rightarrow \infty}{n/n+1}\] \[n/n(1+1/n)=1/(1+1/n)\] take limit to infinity \[1/(1+0)=1\] \[\alpha*1=alpha\]
 one year ago

Fazeelayaz Group TitleBest ResponseYou've already chosen the best response.0
the last term is alpha
 one year ago

Fazeelayaz Group TitleBest ResponseYou've already chosen the best response.0
the serries os monotone
 one year ago

Fazeelayaz Group TitleBest ResponseYou've already chosen the best response.0
the serries is also bounded below becaise it says n>1 soo first term is 1
 one year ago

Fazeelayaz Group TitleBest ResponseYou've already chosen the best response.0
plz give me medal if i helped
 one year ago
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