Question

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

Answer 1

"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.

\ Brought to you by dansmath / (You're welcome!)