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

Consider the series ∑_(𝑛 = 1) ^(∞) (−1)^(𝑛)/(𝑛² + 1). Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

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### Video Transcript

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

Let’s firstly recall that if we have a series the sum of 𝑎𝑛, this is called absolutely convergent if the sum of absolute values converges. It’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 it’s not absolutely convergent, we’ll have to test whether the series itself is convergent. So the first thing we’re going to do is test whether the sum from 𝑛 equals one to ∞ of the absolute value of negative one to the 𝑛th power over 𝑛 squared add one is convergent or divergent.

We first of all notice that negative one to the 𝑛th 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 𝑛th power, this is always going to give us one. Also, note that as 𝑛 goes through positive numbers, 𝑛 squared add one is always going to be positive. So the absolute value of negative one to the 𝑛th power over 𝑛 squared add one can be written as one over 𝑛 squared add one.

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

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

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

Therefore, by the comparison test, the sum from 𝑛 equals one to ∞ of one over 𝑛 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 𝑛 equals one to ∞ of negative one to the 𝑛th power over 𝑛 squared add one is called absolutely convergent.