In this video I go over the infinite series of the form n to the power of -sin 1 and show that is just a divergent p-series. The term nsin 1 can be rewritten as 1/nsin 1 which is of the form 1/np where p is less than 1. We can plug sin 1 into the calculator, in both radians or degrees, and the value is going to be less than 1. Thus, we have a simple divergent p-series since p is less than 1. Hence the answer to the question is it is False, that is, it is NOT a convergent series.
Timestamps:
- Question 2: 0:00
- Solution: False: 0:18
- Calculation check: 1:24
- Divergent p-series with p is less than 1: 2:34
Full video below:
- Infinite Sequences and Series: Review and True-False Quiz:
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