### Video Transcript

Using the πth 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 πth term test for
divergence says that if the limit as π approaches infinity of ππ does not exist
or if the limit is not equal to zero, then the series the sum of ππ from π equals
one to infinity is divergent. We also recall that if the limit is
equal to zero, we canβt 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 doesnβt really
matter. Weβre looking for absolute
divergence, essentially whatβs 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, weβre going to
let ππ be equal to π over π squared plus one. And weβre 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 itβs
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.

Weβre 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 thatβs 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 ππ is equal to zero, we see that the test fails.