### Video Transcript

Consider the series the sum from π
equals one to β of negative one to the πth power over π squared add one. Determine whether the series is
absolutely convergent, conditionally convergent, or divergent.

Letβs firstly recall that if we
have a series the sum of ππ, this is called absolutely convergent if the sum of
absolute values converges. Itβs called conditionally
convergent if the series of absolute values diverges but the series itself still
converges. Otherwise, we say that the series
diverges.

Remember that a series converges if
the partial sums approach a specific limit. So we first of all test whether our
series is absolutely convergent. If itβs not absolutely convergent,
weβll have to test whether the series itself is convergent. So the first thing weβre going to
do is test whether the sum from π equals one to β of the absolute value of negative
one to the πth power over π squared add one is convergent or divergent.

We first of all notice that
negative one to the πth power will always be one or negative one, depending on
whether the power is even or odd. But when we take the absolute value
of negative one to the πth power, this is always going to give us one. Also, note that as π goes through
positive numbers, π squared add one is always going to be positive. So the absolute value of negative
one to the πth power over π squared add one can be written as one over π squared
add one.

So we need to determine whether the
sum from π equals one to β of one over π squared add one is convergent or
divergent. To do this, we recall the
comparison test, which tells us that if π π is less than π π for all 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 this is only valid when π π
and π π are positive.

In other words, if we can find a
series which is bigger than one over π squared add one for all values of π which
converges, we can show that our series converges. Well, one over π squared add one
is always less than one over π squared. And this works for all values of
π.

We remember that a π-series is a
series of the form the sum from π equals one to β of one over π raised to the
π-power. And a π-series converges if π is
greater than one and diverges if π is less than or equal to one. So for one over π squared, as two
is greater than one, one over π squared converges.

Therefore, by the comparison test,
the sum from π equals one to β of one over π squared add one converges. And so because we found the sum of
absolute values of our series to be convergent. Then the original series the sum
from π equals one to β of negative one to the πth power over π squared add one is
called absolutely convergent.