Determine if `sum_(n=1)^(oo) 1/(2n+1)` converges or diverges.

We use the limit comparison test:

Suppose `a_n>0,b_n>0` and `lim_(n->oo)((a_n)/(b_n))=L` with `L` finite and positive, then the two series `sum a_n,sum b_n` either both converge or both diverge.

We know that `sum 1/n` diverges; consider the following limit:

`lim_(n->oo)((1/(2n+1))/(1/n))`

`=lim_(n->oo)(1/(2n+1)*n/1)`

`=lim_(n->oo)(n/(2n+1))`

`=1/2`

**Since the limit is positive and finite, and 1/n diverges, both series must diverge.**

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This is true of the general harmonic `sum_(n=1)^(oo)1/(an+b)` also; it will diverge.

For the given series, need to use the comparison test to show that this series diverges.

`sum_{n=0}^infty 1/{2n+1}`

If we consider each term as a rectangle with height `a_n=1/{2n+1}`, and width 1 then the sum of all these terms is greater than the integral `int_1^infty 1/{2x+1}dx`

`=1/2ln(2x+1)_1^infty`

`=infty`

Now the integral comparison test states that if all the terms are positive and greater than a divergent integral then the series also diverges

**The series diverges.**

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