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`sum_(n=1)^oo 1/(3n-2)` Determine the convergence or divergence of the series.

To evaluate the series `sum_(n=1)^oo 1/(3n-2)` , we may apply Direct Comparison test.


Direct Comparison test is applicable when `sum a_n` and` sum b_n ` are both positive series for all n where `a_n lt=b_n` .


If `sum b_n` converges then `sum a_n` converges.


If `sum a_n` diverges so does the `sum b_n` diverges.


Let `b_n=1/(3n-2)` and `a_n =1/(3n)` since


It follows that `a_n < b_n`


Apply `sum c* a_n = c sum a_n` , we get:


`sum_(n=1)^oo 1/(3n) =1/3sum_(n=1)^oo 1/(n)`


 Apply he p-series test: `sum_(n=1)^oo 1/n^p` is convergent if `pgt1` and divergent if `plt=1` .


For the `sum_(n=1)^oo 1/n ` or `sum_(n=1)^oo 1/n^1` , we have the corresponding value `p=1` . It satisfies the condition `plt=1`  then the series  `sum_(n=1)^oo 1/n` diverges. Therefore, the  `sum_(n=1)^oo 1/(3n)` is a divergent series.


When the `sum a_n = sum_(n=1)^oo 1/(3n) ` is a divergent series then `sum b_n= sum_(n=1)^oo 1/(3n-2)` is also a divergent series.

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