The lecture was excellent, and the teaching efforts (pedagogy) are highly appreciated. If the summation starts from n=1, then, the series converges to the sum S=a*r/(1-r); in case if the summation starts from n=0, then, the series will converge to the sum S=a/(1-r). In any case, we have a quantity "a" but not "a1", and where a=1. The confusion at the end came because of writing "a1", but this is not to be considered as a mistake. Regarding degrees or radians: Since the professor spelled out the word degrees, we can use 400 degrees to evaluate numerical value. In case if nothing is specified about 400, then radian must be used for finding numerical value.
![Comparison Tests [ Direct and Limit]](http://i.ytimg.com/vi/ITAUERlNDrU/mqdefault.jpg)
![Comparison Tests [ Direct and Limit]](/img/tr.png)
The lecture was excellent, and the teaching efforts (pedagogy) are highly appreciated.
If the summation starts from n=1, then, the series converges to the sum S=a*r/(1-r); in case if the summation starts from n=0, then, the series will converge to the sum S=a/(1-r). In any case, we have a quantity "a" but not "a1", and where a=1. The confusion at the end came because of writing "a1", but this is not to be considered as a mistake.
Regarding degrees or radians: Since the professor spelled out the word degrees, we can use 400 degrees to evaluate numerical value. In case if nothing is specified about 400, then radian must be used for finding numerical value.
I teach my students to think "first term of the series" instead of a_0 or a_1 or whatever.
i'm gonna take the final-term exam and this vid help me a lot, thanks professor
Make a video for alternating series test
nice
Strictly speaking, here, 400 radians : |sin 400| = |sin(37,9° appr)| < 1; no?
He does mention that its 400 degrees, not radians but yeah he didnt write it that way