# Question Video: Using the πth-Term Test to Determine Whether a Series Is Divergent and Determining When the Test Fails Mathematics • Higher Education

Using the πth term test, determine whether the series β_(π = 0) ^(β) π/(πΒ² + 1) is divergent or the test fails.

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### Video Transcript

Using the πth term test, determine whether the series the sum of π over π squared plus one from π equals zero to infinity is divergent or the test fails.

Remember, the πth term test for divergence says that if the limit as π approaches infinity of ππ does not exist or if the limit is not equal to zero, then the series the sum of ππ from π equals one to infinity is divergent. We also recall that if the limit is equal to zero, we canβt tell whether the series converges or diverges and we say that the test fails.

Now notice the sum in our question is from π equals zero to infinity, as opposed to from π equals one to infinity. In practice, this doesnβt really matter. Weβre looking for absolute divergence, essentially whatβs happening to the series as π gets larger and larger. And if we look carefully, we see that when π equals zero, our first term is also zero. So we can split this into zero plus the sum from π equals one.

In our case then, weβre going to let ππ be equal to π over π squared plus one. And weβre going to evaluate the limit as π approaches infinity of π over π 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 π 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 π over π squared plus one. We do so by dividing both the numerator and denominator by π squared.

Remember, we can do this as it creates an equivalent fraction. And we choose π squared as itβs the highest power of π in our denominator. So we get the limit as π approaches infinity of π over π squared over π squared over π squared plus one over π squared. This simplifies to the limit as π approaches infinity of one over π over one plus one over π squared.

Weβre then going to use the division law for limits. This says that the limit as π₯ approaches π of π of π₯ over π of π₯ is equal to the limit as π₯ approaches π of π of π₯ over the limit as π₯ approaches π of π of π₯. And thatβs provided the limits exist and the limit of π of π₯ is not equal to zero. And so this becomes the limit as π approaches infinity of one over π over the limit as π approaches infinity of one plus one over π squared.

And we can now substitute π equals infinity into this expression. As π gets larger, one over π gets smaller. Ultimately, it approaches zero. Similarly, as π gets larger, one over π 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 π approaches infinity of ππ is equal to zero, we see that the test fails.