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Test the series for convergence or divergence. [infinity] ∑ sin(9n) / 1+9^n n=1

User Virullius
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1 Answer

3 votes

Answer:

Converges by Direct Comparison Test

Explanation:

For the infinite series
\displaystyle \sum^\infty_(n=1)(\sin(9n))/(1+9^n), we can use the direct comparison test. We need to check for absolute convergence, so let's assume
\displaystyle \sum^\infty_(n=1)\biggr|(\sin(9n))/(1+9^n)\biggr|\leq\sum^\infty_(n=1)(1)/(1+9^n). Since
\displaystyle \sum^\infty_(n=1)(1)/(9^n) is a geometric series with
\displaystyle r=(1)/(9) < 1, then that series converges. This implies that
\displaystyle \sum^\infty_(n=1)(1)/(1+9^n) converges, and so
\displaystyle \sum^\infty_(n=1)(\sin(9n))/(1+9^n) converges by the direct comparison test.

User Husterknupp
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