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El_Arrow
 one year ago
determine whether the series is convergent or divergent
El_Arrow
 one year ago
determine whether the series is convergent or divergent

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El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty} ((1/n) + (1/n+1))\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, the easiest test for convergence/divergence of a series is to take the limit of n approaches infinity.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Given a series, \[\sum_{n=1}^{\infty} a_{n}\]. If \[\lim_{n \rightarrow \infty} a_{n} = 0 \], it is convergent. Otherwise, it is divergent.

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0so you use the integral test for this one or no? @math1234

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436734133011:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is the limit in this case?

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0from t to 1 @math1234

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0but how do you find the limit of that?

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The limit of the first term is 0 right? Same as the second term.

El_Arrow
 one year ago
Best ResponseYou've already chosen the best response.0so it completely diverges?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0That is not what the theorem says, it says that if the limit DOES not converge to 0, then the series does not converge. It does NOT say that if the limit converges to 0 then the series converges. Every horse has four legs, but not every four legged animal is a horse

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0integral test is a good idea

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this does not converge, obviously:$$\frac1{n+1}\ge 0\implies \frac1{n}\le\frac1{n}+\frac1{n+1}\implies \sum_{n=1}^N\frac1n\le \sum_{n=1}^N\left(\frac1n+\frac1{n+1}\right)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and yet we know that \(\sum_{n=1}^\infty\frac1n=\lim_{N\to\infty}\sum_{n=1}^N\frac1n\to\infty\) as the harmonic series diverges, and surely something bigger than the harmonic series diverges, too
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