### Video Transcript

A series, the sum from π equals zero to β of the sequence π π, satisfies that the limit as π approaches β of the absolute value of π at π plus one divided by π π is equal to one. What can you conclude about the convergence of the series?

Weβre asked what we can conclude about the convergence of this series given that the limit as π approaches β of the absolute value of the quotient of successive terms is equal to one. Since weβre given information about the limit as π approaches β of the absolute value of the ratio of successive terms, we could consider what the ratio test will tell us about the series. From the ratio test, we know if this limit was less than one, then our series would converge. And if our limit was greater than one, then our series would diverge. However, the ratio test tells us that if this limit is equal to one, then the test will be inconclusive.

To help see why the ratio test is inconclusive in this case, letβs consider the following three series. First, the sum from π equals zero to β of just a constant one. Second, the sum from π equals zero to β of one divided by π plus one all squared. And third, the sum from π equals zero to β of negative one to the πth power divided by π plus one. Letβs consider the limit as π approaches β of the absolute value of the ratio of successive terms in our first series. Our sequence is just equal to the constant one. So π π plus one is just equal to one. And π π is just equal to one. So this gives us the limit as π approaches β of the absolute value of one divided by one. And the absolute value of one divided by one is just equal to one. And the limit of a constant is just equal to that constant itself. So this is equal to one. There are several different ways to tell that this series is divergent. For example, itβs πth partial sum is equal to π, which is unbounded. Or we can consider this series is as a π series, with π equal to zero.

Letβs now consider the limit as π approaches β of the absolute value of the ratio of successive terms in our second series. Itβs worth noting at this point that since π is approaching β, we can reduce the indices in our limit by one to make the algebra a little easier. This gives us that our numerator π π is equal to one divided by π plus one all squared. And our denominator π π minus one is equal to one divided by π squared. We can simplify our fraction being divided by another fraction by flipping the fraction in our denominator over and multiply. Doing this gives us π squared divided by π plus one all squared. Next, if we distribute our exponent over our parentheses, we get π squared plus two π plus one. Giving us the limit as π approaches β of the absolute value of π squared divided by π squared plus two π plus one.

There are many different ways of evaluating limits such as this. One such way is to divide both the numerator and the denominator by π squared. Dividing the numerator by π squared gives us π squared divided by π squared, which is just equal to one. And dividing the denominator by π squared gives us π squared plus two π plus one all over π squared. Which is just equal to one plus two over π plus one over π squared. This then gives us the limit as π approaches β of the absolute value of one divided by one plus two over π plus one over π squared. Since π is approaching β, two over π and one over π squared are both approaching zero. Therefore, we can evaluate our limit as the absolute value of one divided by one, which is just equal to one. To determine the convergence or divergence of our second series, letβs note that our second series is actually equal to a π series, with π equal to two. Therefore, we can conclude that our second series is convergent because itβs a π series, with π equal to two. In fact, since one over π squared is always positive, we can conclude that our second series is absolutely convergent.

Finally, letβs consider the limit as π approaches β of the absolute value of the ratio of successive terms in our third series. Since π is approaching β, we can reduce the indices in our limit by one to make the algebra a little bit easier. So our numerator is π π, which is negative one to the πth power divided by π plus one. And our denominator is π π minus one, which is negative one to the power of π minus one divided by π. We can simplify this by flipping the fraction in our denominator and multiplying. We can then cancel out the shared factors of negative one in the numerator and the denominator. To give us the limit as π approaches β of the absolute value of negative π divided by π plus one.

We can evaluate this limit similarly to how we did in the second series. Weβll divide the numerator and the denominator by π. Dividing the numerator of negative π by π gives us negative one. And dividing the denominator of π plus one by π gives us one plus one over π. This gives us the limit as π approaches β of the absolute value of negative one divided by one plus one over π. Since π is approaching β, one divided by π is approaching zero. And we know the absolute value of negative one divided by one is just equal to one. So we can evaluate our limit to just be equal to one.

Next, if we notice that one divided by π plus one is a positive decreasing sequence, then the sum from π equals zero to β of negative one to the πth power divided by π plus one must be a convergent series by the alternating series test. However, if we were to consider the sum from π equals zero to β of the absolute value of negative one to the πth power divided by π plus one. We see that this series is equal to the sum from π equals one to β of one divided by π, also known as the harmonic series. Since the harmonic series is divergent and this is equal to the absolute value of this series. We can conclude that our third series is conditionally convergent.

So what we have done is we have found three different series, each of which has the limit as π approaches β of the absolute value of the ratio of successive terms equal to one. Where one series is divergent. One series is absolutely convergent. And the third one is conditionally convergent. Therefore, since our series can be divergent, absolutely convergent, or conditionally convergent, we must know that our test was inconclusive.