El_Arrow
  • El_Arrow
determine whether the series is convergent or divergent
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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El_Arrow
  • El_Arrow
\[\sum_{n=1}^{\infty} ((1/n) + (1/n+1))\]
El_Arrow
  • El_Arrow
@SolomonZelman
El_Arrow
  • El_Arrow
@campbell_st

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More answers

anonymous
  • anonymous
Well, the easiest test for convergence/divergence of a series is to take the limit of n approaches infinity.
anonymous
  • anonymous
Given 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
  • El_Arrow
so you use the integral test for this one or no? @math1234
El_Arrow
  • El_Arrow
|dw:1436734133011:dw|
El_Arrow
  • El_Arrow
like that?
anonymous
  • anonymous
Correct.
anonymous
  • anonymous
What is the limit in this case?
El_Arrow
  • El_Arrow
from t to 1 @math1234
El_Arrow
  • El_Arrow
but how do you find the limit of that?
anonymous
  • anonymous
\[\lim_{n \rightarrow \infty} \frac{ 1 }{ n } + \frac{ 1 }{ n + 1}\]
anonymous
  • anonymous
The limit of the first term is 0 right? Same as the second term.
El_Arrow
  • El_Arrow
so it completely diverges?
zzr0ck3r
  • zzr0ck3r
That 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
  • zzr0ck3r
integral test is a good idea
anonymous
  • anonymous
this 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
  • anonymous
and 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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