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anonymous
 3 years ago
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
anonymous
 3 years ago
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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i would try with \(\alpha =1\) first

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in 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 \(\lim_{n\to \infty}\frac{n}{n+1}=1\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i meant "bounded above by one" not below!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it is monotone increasing as you can check by taking the derivative and seeing that is always positive

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then try the general case with \(\alpha\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes 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 \[1/(1+0)=1\] \[\alpha*1=alpha\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the last term is alpha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the serries os monotone

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the serries is also bounded below becaise it says n>1 soo first term is 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0plz give me medal if i helped
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