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## Comparison tests for convergence

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# Worked example: direct comparison test

AP Calc: LIM‑7 (EU), LIM‑7.A (LO), LIM‑7.A.8 (EK)

## Video transcript

- [Voiceover] Let's think
about the infinite series, so we're going to go from
n equals one to infinity, of one over two to the n plus n. And what I want to do
is see if we can prove whether this thing converges or diverges. And as you can imagine based
on the context of where this video shows up on
Khan Academy that maybe we will do it using the comparison test. At any point if you feel
like you can kinda take this to the finish line, feel free to pause the video and do so. So in order to kind of
figure out or get a sense for this series right over
here, it never hurts to kind of expand it out a
little bit, so let's do that. So this would be equal
to when n equals one this is gonna be one over
two to the one plus one, so it's gonna be one over two plus one, it's gonna be one third
plus that's n equals one. When n equals two it's gonna
be one over two squared, which is four plus two plus one over six. Plus let's see we go
to three, n equals one, n equals two, n equals
three is gonna be one over two to the third, which
is eight plus three is 11. So one over 11, maybe
I'll do one more term. Two the fourth power is going
to be 16 plus four is 20. Plus one over 20 and obviously we just keep going on and on and on. So it looks, it feels like
this thing could converge. All of our terms are
positive, but they are getting smaller and smaller quite fast. And if we really look at
the behavior of the terms as n gets larger and
larger, we see that the two to the n in the denominator will grow much, much faster than the n will. So this kind of behaves
like one over two to the n, which is a clue of something that we might be able to use for the comparison test. So let me just write that down. So we have one over, so
the infinite series from n equals one to infinity
of one over two to the n, and so when n equals one this is going to be equal to one half. When n is equal to two this is going to be equal to one fourth. When n is equal to three
this is equal to one eighth. When n is equal to four this
is equal to one sixteenth. And we go on and on and on and on. And what's interesting about
this is we recognize it. This is a geometric
series, so let me be clear. This thing right over here,
that is the same thing as the sum of from n
equals one to infinity of one half to the n power, just
writing it in a different way. And since the absolute
value of one half, which is just one half, so
because the absolute value of one half is less than one we know that this geometric series converges,
we know that it converges And actually we even
have formulas for finding the exact sum of or to figure
out what it converges to. And so we know this thing
converges and we see that actually these two series
combined meet all of the constraints we need for
the comparison test. So let's go back to what we
wrote about the comparison test. So the comparison test,
we have two series, all of their terms are greater
than or equal to zero. All of these terms are
greater than or equal to zero. And then for the corresponding
terms in one series, all of them are going
to be less than or equal to the corresponding
terms in the next one. And so if we look over here,
we can consider this one, the magenta series, this
is kind of our infinite series of dealing with A
sub n and that this right over here is, well I
already did it in blue this is kind of the blue series. And notice all their
terms are nonnegative and the corresponding terms
one half is greater than one third, one fourth is
greater than one sixth, one eighth is greater than one eleventh. One over two to the n is
always going to be greater than one over two to the n plus n for the n that we care about here. And since we know that
this converges, since we know that the larger one converges. It's a geometric series
where the common ratio the absolute value of
the common ratio is less than one, since we know
that the larger series converges therefore, the
smaller series or the one where every corresponding
term is less than the one in the blue one,
that one must also converge. So by the comparison test the series in question must also converge.

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