How do you use the direct Comparison test on the infinite series sum_(n=1)^oo(1+sin(n))/(5^n) ?

1 Answer
dansmath
Aug 18, 2014

"Using a test" means deciding if the series converges or diverges. If you can compare this series to a larger one that converges, that means this series converges too.

The value of sin(x) is always between -1 and 1; sin(n) >= -1 means our series is all non-negative terms. And since we have sin(n) <= 1 for all n, comparing the nth term we get:

(1+sin(n))/(5^n) < 2/(5^n)

This means our sum is less than sum_(n=1)^(oo)2/(5^n), which converges as a geometric series with r = 1/5 < 1.

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