determine whether the series is convergent or divergent

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determine whether the series is convergent or divergent

Mathematics
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\[\sum_{n=1}^{\infty} ((1/n) + (1/n+1))\]

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Well, the easiest test for convergence/divergence of a series is to take the limit of n approaches infinity.
Given a series, \[\sum_{n=1}^{\infty} a_{n}\]. If \[\lim_{n \rightarrow \infty} a_{n} = 0 \], it is convergent. Otherwise, it is divergent.
so you use the integral test for this one or no? @math1234
|dw:1436734133011:dw|
like that?
Correct.
What is the limit in this case?
from t to 1 @math1234
but how do you find the limit of that?
\[\lim_{n \rightarrow \infty} \frac{ 1 }{ n } + \frac{ 1 }{ n + 1}\]
The limit of the first term is 0 right? Same as the second term.
so it completely diverges?
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
integral test is a good idea
this does not converge, obviously:$$\frac1{n+1}\ge 0\implies \frac1{n}\le\frac1{n}+\frac1{n+1}\implies \sum_{n=1}^N\frac1n\le \sum_{n=1}^N\left(\frac1n+\frac1{n+1}\right)$$
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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