`sum_(n=1)^oo 1/sqrt(n^3+1)` Use the Direct Comparison Test to determine the convergence or divergence of the series.

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The 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 the series is `sum_(n=1)^oo1/sqrt(n^3+1)`

Let `a_n=1/sqrt(n^3+1)` and `b_n=1/sqrt(n^3)=1/n^(3/2)`

`1/sqrt(n^3)>1/sqrt(n^3+1)>0`  for `n>=1`

The series `sum_(n=1)^oo1/n^(3/2)` is a p-series with p=`3/2`

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The 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 the series is `sum_(n=1)^oo1/sqrt(n^3+1)`

Let `a_n=1/sqrt(n^3+1)` and `b_n=1/sqrt(n^3)=1/n^(3/2)`

`1/sqrt(n^3)>1/sqrt(n^3+1)>0`  for `n>=1`

The series `sum_(n=1)^oo1/n^(3/2)` is a p-series with p=`3/2`

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

As per the p-series test the series `sum_(n=1)^oo1/sqrt(n^3)` converges, so the series `sum_(n=1)^oo1/sqrt(n^3+1)` as well, converges by the direct comparison test.

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