### Video Transcript

State whether the series the sum
from π equals one to β of negative one to the power of π add one multiplied by two
over the square root of π add one converges absolutely, conditionally, or not at
all.

Firstly, recall that for a series
π π, this is absolutely convergent if the series of absolute values converges. And itβs conditionally convergent
if the series of absolute values diverges. But the series itself
converges. So letβs first of all find out
whether this series is absolutely convergent or not. This means testing whether the
series from π equals one to β of the absolute value of negative one to the power of
π add one multiplied by two over the square root of π add one is convergent or
divergent.

Well, negative one to the power of
π add one is always going to be one or negative one. But if we take the absolute value,
it will always be one. Whereas two over the square root of
π add one will always be positive because π runs through positive values. So we can write this as the sum
from π equals one to β of two over the square root of π add one. Then, we can use the constant
multiplication rule to bring the two to the front of the sum. From here, we need to work out
whether this series converges or diverges. One way we can actually do this is
with a direct comparison with the harmonic series.

Because for π is greater than two,
we have that one over π is less than one over the square root of π add one. And we know that if we have π π
less than π π where π π diverges, then π π also diverges. And we know that the sum from π
equals one to β of one over π is the harmonic series which diverges. Then, the sum from π equals one to
β of one over the square root of π add one also diverges. So we found that the series of
absolute values diverges, which means that this series is not absolutely
convergent. But it could still be conditionally
convergent. So weβre going to test the series
itself for convergence. So letβs clear some space.

We can firstly bring the constant
two to the front of the sum. And then, if we look at this
negative one to the power of π add one, this creates an alternating series because
it makes the terms alternate between positive and negative. So we can decide whether this
series is convergent or divergent using the alternating series test. Remember that this says for a
series π π, where π π is equal to negative one to the power of π add one
multiplied by π π, if π π is decreasing and the limit as π approaches β of π
π is equal to zero, then the series π π is convergent. So for our series π π is equal to
one over the square root of π add one. But is this decreasing? Well, for π π to be decreasing,
we need π π to be greater than π π add one.

Well, we can see that as π
increases by one, the square root of π add one is going to get bigger. So one over the square root of π
add one is going to decrease. So π π is decreasing. Then, we need to check whether the
limit as π approaches β of π π is equal to zero. In other words, is the limit as π
approaches β of one over the square root of π add one zero? Well, the square root of π add one
is increasing as π gets bigger. So this will be one over β. So as π approaches β, then one
over the square root of π add one approaches one over β. And so the limit as π approaches β
is zero. So both of these conditions are
satisfied. So this series is convergent.

Remember that we said that a series
is conditionally convergent if the series of absolute values diverges but the series
converges. And thatβs exactly what weβve found
here. The series of absolute values was
divergent. But we found the series itself to
be convergent. So we can conclude that this series
converges conditionally.