`sum_(n=1)^oo 1/(2n-1)` Use the Direct Comparison Test to determine the convergence or divergence of the series.

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Direct comparison test is applicable when `suma_n` and`sumb_n` are both positive sequences for all n, such that `a_n<=b_n` .It follows that:

If `sumb_n` converges then `suma_n` converges.

If `suma_n` diverges then `sumb_n` diverges.

`sum_(n=1)^oo1/(2n-1)`

Let `b_n=1/(2n-1)` and `a_n=1/(2n)`

`1/(2n-1)>1/(2n)>0`   for `n>=1`

As per p series test `sum_(n=1)^oo1/n^p` is convergent if `p>1` and...

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Direct comparison test is applicable when `suma_n` and`sumb_n` are both positive sequences for all n, such that `a_n<=b_n` .It follows that:

If `sumb_n` converges then `suma_n` converges.

If `suma_n` diverges then `sumb_n` diverges.

`sum_(n=1)^oo1/(2n-1)`

Let `b_n=1/(2n-1)` and `a_n=1/(2n)`

`1/(2n-1)>1/(2n)>0`   for `n>=1`

As per p series test `sum_(n=1)^oo1/n^p` is convergent if `p>1` and divergent if `p<=1`

`sum_(n=1)^oo1/(2n)=1/2sum_(n=1)^oo1/n`

`sum_(n=1)^oo1/n`  is a p-series with p=1, so it diverges.

Since `sum_(n=1)^oo1/(2n)` diverges ,the series `sum_(n=1)^oo1/(2n-1)` diverges too by the direct comparison test. 

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