Video: Determining Whether a Given Series Is Convergent or Divergent Using the Comparison Test

Use the comparison test to determine whether the series βˆ‘_(𝑛 = 1)^(∞) 1/(𝑛 + 2) is convergent or divergent.

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Video Transcript

Use the comparison test to determine whether the series the sum from 𝑛 equals one to ∞ of one over 𝑛 add two is convergent or divergent.

Let’s begin this question by writing out the comparison test. The comparison test tells us that for two series the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛, where π‘Ž 𝑛 and 𝑏 𝑛 are both greater than or equal to zero for all 𝑛. Then if the series 𝑏 𝑛 converges and π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all 𝑛, then the series π‘Ž 𝑛 converges. If the series 𝑏 𝑛 diverges and π‘Ž 𝑛 is greater than or equal to 𝑏 𝑛 for all 𝑛, then the series π‘Ž 𝑛 diverges.

If we work out the first few terms of our sum, we find that this is equal to one over three add one over four add one over five add one over six and so on. We can see some similarities here with the harmonic series, which we know to be a divergent series. So we might suspect that our series is divergent. By the comparison test, if we can find another divergent series which is smaller than our series, we can conclude that our series diverges.

Unfortunately, we cannot compare it with the harmonic series because one over 𝑛 is not less than or equal to one over 𝑛 add two. But what if we compare it instead with the series the sum from 𝑛 equals one to ∞ of one over three 𝑛. Well, one over 𝑛 add two is greater than or equal to one over three 𝑛 for all values of 𝑛. So if the series the sum from 𝑛 equals one to ∞ of one over three 𝑛 diverges, we’ve proved that our series diverges.

Well, we can start by rewriting the series the sum from 𝑛 equals one to ∞ of one over three 𝑛 as one over three multiplied by the sum from 𝑛 equals one to ∞ of one over 𝑛. And now, we see that the sum from 𝑛 equals one to ∞ of one over 𝑛 is just the harmonic series which we know is divergent.

And as a general result, the constant doesn’t affect the divergence of the series. So the sum from 𝑛 equals one to ∞ of one over three 𝑛 is also divergent. So as we’ve managed to find 𝑏 𝑛, which is less than or equal to our π‘Ž 𝑛, and the series 𝑏 𝑛 is divergent, then our series is also divergent by the comparison test.

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