### Video Transcript

Consider the series the sum from π
equals one to β of sin of π over π cubed. Determine whether the series is
absolutely convergent, conditionally convergent, or divergent.

Recall that a series π π is
absolutely convergent if the series of absolute values is convergent. And if we find that the series is
not absolutely convergent, it may still be conditionally convergent. So we then test the series for
convergence or divergence. So letβs begin by testing this
series for absolute convergence. So we want to find out whether the
sum from π equals one to β of the absolute value of sin of π over π cubed is
convergent or divergent.

Well, because π only runs through
positive values from one to β, π cubed is always going to be positive. So this is just the sum from π
equals one to β of the absolute value of sin of π over π cubed. Now we know that sin of π will
always be between negative one and one. So we can say that the absolute
value of sin of π will always be less than or equal to one, which means that we can
write the absolute value of sin of π over π cubed is less than or equal to one
over π cubed. Writing it this way allows us to do
a direct comparison. Recall that this means if π π is
less than π π and the sum from π equals one to β of π π converges, then the sum
from π equals one to β of π π also converges.

And one over π cubed is actually a
series we recognize. Recall that a π-series is a series
of the form the sum for π equals one to β of one over π to the π power. And this converges if π is greater
than one and diverges if π is less than or equal to one. So one over π cubed is a π-series
with π equal to three. So one over π cubed converges. So by direct comparison, the
absolute value of sin of π over π cubed also converges. Then, because we found the series
of absolute values to be convergent, then our series the sum from π equals one to β
of sin of π over π cubed is absolutely convergent.