`sum_(n=1)^oo 1/(4root(3)(n)-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 series for all n such that `a_n<=b_n`

If `sumb_n` converges , then `suma_n` converges

If `suma_n` diverges , then `sumb_n` diverges.

Given series is `sum_(n=1)^oo1/(4root(3)(n)-1)`

Let `b_n=1/(4root(3)(n)-1)` and `a_n=1/(4root(3)(n))=1/4(1/n^(1/3))`

`1/(4root(3)(n)-1)>1/(4root(3)(n))>0`  for `n>=1`

The series `sum_(n=1)^oo1/4(1/n^(1/3))` is a p-series with p=`1/3<1`

The p-series...

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Direct comparison test is applicable when `suma_n` and `sumb_n` are both positive series for all n such that `a_n<=b_n`

If `sumb_n` converges , then `suma_n` converges

If `suma_n` diverges , then `sumb_n` diverges.

Given series is `sum_(n=1)^oo1/(4root(3)(n)-1)`

Let `b_n=1/(4root(3)(n)-1)` and `a_n=1/(4root(3)(n))=1/4(1/n^(1/3))`

`1/(4root(3)(n)-1)>1/(4root(3)(n))>0`  for `n>=1`

The series `sum_(n=1)^oo1/4(1/n^(1/3))` is a p-series with p=`1/3<1`

The p-series `sum_(n=1)^oo1/n^p` converges if `p>1` and diverges if `0<p<=1`

Since the series diverges `sum_(n=1)^oo1/(4root(3)(n))` as per the p-series test and so the series `sum_(n=1)^oo1/(4root(3)(n)-1)` as well , diverges by the direct comparison test.

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