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

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

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

`1/(sqrt(n)-1)>1/sqrt(n)>0` for `n>=2`

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

The p-series `sum_(n=1)^oo1/n^p` , is convergent if `p>1` and divergent if `0<p<=1`

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

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

`1/(sqrt(n)-1)>1/sqrt(n)>0` for `n>=2`

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

The p-series `sum_(n=1)^oo1/n^p` , is convergent if `p>1` and divergent if `0<p<=1`

For the series `sum_(n=2)^oo1/n^(1/2)` p=`1/2<1` so it diverges as per the p-series test.

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

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