How do you use the direct Comparison test on the infinite series $sum_(n=1)^oo9^n/(3+10^n)$ ?

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Answer 1 · 2014-09-11T17:12:10

By Comparison Test, we can conclude that the series
$sum_{n=1}^infty{9^n}/{3+10^n}$ converges.

Let us look at some details.
For all $n geq 1$ ,
$9^n/{3+10^n} leq 9^n/10^n=(9/10)^n$

By Geometric Series Test,
$sum_{n=1}^infty(9/10)^n$ converges since $|r|=9/10<1$

Hence, by Comparison Test,
$sum_{n=1}^infty{9^n}/{3+10^n}$ also converges.

How do you use the direct Comparison test on the infinite series sum_(n=1)^oo9^n/(3+10^n)sum_(n=1)^oo9^n/(3+10^n) ?