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

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1. El_Arrow

$\sum_{n=1}^{\infty} ((1/n) + (1/n+1))$

2. El_Arrow

@SolomonZelman

3. El_Arrow

@campbell_st

4. anonymous

Well, the easiest test for convergence/divergence of a series is to take the limit of n approaches infinity.

5. 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.

6. El_Arrow

so you use the integral test for this one or no? @math1234

7. El_Arrow

|dw:1436734133011:dw|

8. El_Arrow

like that?

9. anonymous

Correct.

10. anonymous

What is the limit in this case?

11. El_Arrow

from t to 1 @math1234

12. El_Arrow

but how do you find the limit of that?

13. anonymous

$\lim_{n \rightarrow \infty} \frac{ 1 }{ n } + \frac{ 1 }{ n + 1}$

14. anonymous

The limit of the first term is 0 right? Same as the second term.

15. El_Arrow

so it completely diverges?

16. 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

17. zzr0ck3r

integral test is a good idea

18. anonymous

this does not converge, obviously:$$\frac1{n+1}\ge 0\implies \frac1{n}\le\frac1{n}+\frac1{n+1}\implies \sum_{n=1}^N\frac1n\le \sum_{n=1}^N\left(\frac1n+\frac1{n+1}\right)$$

19. 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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