# Video: Determining Whether a Given Series Is Absolutely Convergent, Conditionally Convergent, or Divergent

Consider the series β_(π = 1) ^(β) (β1)^(π)/(πΒ² + 1). Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

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### Video Transcript

Consider the series the sum from π equals one to β of negative one to the πth power over π squared add one. Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

Letβs firstly recall that if we have a series the sum of ππ, this is called absolutely convergent if the sum of absolute values converges. Itβs 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 itβs not absolutely convergent, weβll have to test whether the series itself is convergent. So the first thing weβre going to do is test whether the sum from π equals one to β of the absolute value of negative one to the πth power over π squared add one is convergent or divergent.

We first of all notice that negative one to the πth 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 πth 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 πth 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 πth power over π squared add one is called absolutely convergent.