Video: Using the 𝑛th-Term Test to Determine Whether a Series Is Divergent and Determining When the Test Fails

Using the 𝑛th term test, determine whether the series βˆ‘_{𝑛 = 0} ^{∞} 𝑛/(𝑛² + 1) is divergent or the test fails.

02:35

Video Transcript

Using the 𝑛th term test, determine whether the series the sum of 𝑛 over 𝑛 squared plus one from 𝑛 equals zero to infinity is divergent or the test fails.

Remember, the 𝑛th term test for divergence says that if the limit as 𝑛 approaches infinity of π‘Žπ‘› does not exist or if the limit is not equal to zero, then the series the sum of π‘Žπ‘› from 𝑛 equals one to infinity is divergent. We also recall that if the limit is equal to zero, we can’t tell whether the series converges or diverges and we say that the test fails.

Now notice the sum in our question is from 𝑛 equals zero to infinity, as opposed to from 𝑛 equals one to infinity. In practice, this doesn’t really matter. We’re looking for absolute divergence, essentially what’s happening to the series as 𝑛 gets larger and larger. And if we look carefully, we see that when 𝑛 equals zero, our first term is also zero. So we can split this into zero plus the sum from 𝑛 equals one.

In our case then, we’re going to let π‘Žπ‘› be equal to 𝑛 over 𝑛 squared plus one. And we’re going to evaluate the limit as 𝑛 approaches infinity of 𝑛 over 𝑛 squared plus one. And when evaluating a limit, we should always check whether we can use direct substitution. In this case, if we were to substitute 𝑛 equals infinity into the expression, we get infinity over infinity, which we know to be indeterminate. So instead, we look for a way to manipulate the expression 𝑛 over 𝑛 squared plus one. We do so by dividing both the numerator and denominator by 𝑛 squared.

Remember, we can do this as it creates an equivalent fraction. And we choose 𝑛 squared as it’s the highest power of 𝑛 in our denominator. So we get the limit as 𝑛 approaches infinity of 𝑛 over 𝑛 squared over 𝑛 squared over 𝑛 squared plus one over 𝑛 squared. This simplifies to the limit as 𝑛 approaches infinity of one over 𝑛 over one plus one over 𝑛 squared.

We’re then going to use the division law for limits. This says that the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯. And that’s provided the limits exist and the limit of 𝑔 of π‘₯ is not equal to zero. And so this becomes the limit as 𝑛 approaches infinity of one over 𝑛 over the limit as 𝑛 approaches infinity of one plus one over 𝑛 squared.

And we can now substitute 𝑛 equals infinity into this expression. As 𝑛 gets larger, one over 𝑛 gets smaller. Ultimately, it approaches zero. Similarly, as 𝑛 gets larger, one over 𝑛 squared also approaches zero. And so our limit becomes zero over one plus zero, which is of course simply zero. And so since the limit as 𝑛 approaches infinity of π‘Žπ‘› is equal to zero, we see that the test fails.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.