# `sum_(n=1)^oo 3^n/(2^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

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

`3^n/(2^n-1)>3^n/2^n>0`  for `n>=1`

`sum_(n=1)^oo(3/2)^n` is a geometric series with ratio r=`3/2>1`

A geometric series with `|r|>=1` diverges.

The geometric series `sum_(n=1)^oo(3/2)^n` diverges and so the series `sum_(n=1)^oo3^n/(2^n-1)` diverges as well, by the direct comparison test.

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