`sum_(n=0)^oo 4^n/(5^n+3)` 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 `b_n` converges ,then `a_n` converges.

If `a_n` diverges, then `b_n` diverges.

`sum_(n=0)^oo4^n/(5^n+3)`

Let `a_n=4^n/(5^n+3)` and `b_n=4^n/5^n=(4/5)^n`

`4^n/5^n>4^n/(5^n+3)>0` for `n>=1`

`sum_(n=0)^oo(4/5)^n` is a geometric series with ratio r`=4/5<1`

A geometric series with ratio r , such that `|r|<1` converges.

The geometric series `sum_(n=0)^oo(4/5)^n` converges,so the series `sum_(n=0)^oo4^n/(5^n+3)` converges as well , by the direct comparison test.

Comments

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