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anonymous
 one year ago
Series exercie
anonymous
 one year ago
Series exercie

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty} \frac{ n }{ n ^{2} +1 }\]

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0Okay, what's the objective?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to find if it is convergent,conditonally convergent or divergent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ayeshaafzal221 helped me yesterday with it,I just wanna see another way of solving it

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Hint: use the integral test.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got it like this: \[\lim_{n \rightarrow \infty} \frac{ \frac{ n }{ n ^{2} } }{ \frac{ n ^{2} }{ n ^{2} +1}} = \lim_{n \rightarrow \infty} \frac{ n }{ 2 } =\frac{ \infty }{ 2 }=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is my solution right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0You could also use the comparison test : for \(n\ge 1\), we have \(n^2+1 \le n^2+n^2 = 2n^2\) \(\implies \dfrac{1}{n^2+1}\ge \dfrac{1}{2n^2}\) \(\implies \dfrac{n}{n^2+1}\ge \dfrac{n}{2n^2} = \dfrac{1}{2n}\) Since \(\sum \dfrac{1}{2n}\) doesn't converge as this is a harmonic series, the series \(\sum \dfrac{n}{n^2+1}\) doesn't converge by comparison test

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So my solution is not right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I would give you 1/10 for the attempt because I don't really get what test you have used..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Also why do you think \(\dfrac{\infty}{2}\) is anything close to \(0\) ?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\frac{n}{n^2+1}\ne\frac{n/n^2}{n^2/n^2+1}\) so it is not the equivalent series, in any case \(\inf/2=\inf\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i meant infinite,not 0,sorry

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0do you know what does it mean to 'converge' or 'diverge' ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope,I know if it dose not equal 0,it means it is divergent,if it is 0,then it's complicated

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0before trying the convergence tests, you need to know the meaning of terms "convergent sreies" and "divergent series"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can I use L'Hospital for this exercise?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get 1/2 if I use L'Hospital and it means it is divergent,right?
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