Main content

## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 7

Lesson 10: Ratio & alternating series tests# Ratio test

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.11 (EK)

The ratio test is a most useful test for series convergence. It caries over intuition from geometric series to more general series. Learn more about it here.

## Want to join the conversation?

- You can't prove that n^10/n! doesn't diverge with the divergence test as stated in2:40because lim an = 0 so it can still diverge ou converge. Or am I wrong?(43 votes)
- Absolutely correct, a series such as 1/n (harmonic series) diverges, even though the limit as n goes to infinity leads to 0 for an, so Sal made a mistake.(37 votes)

- So I am currently taking BC Calculus and truly appreciate the help Khan Academy has been giving. I am learning about Series and Sequences and can't seem to find a video on the Root Test. I might not have seen it and would appreciate if anyone can guide me to a video or website that could help me with this test.(34 votes)
- The same conclusions apply as with this test. The only difference is that you take the limit of the absolute value of the nth root of the function.(11 votes)

- Why did you keep (n+1)^10/(n+1) instead of just making (n+1)^10 (n+1)^9?(17 votes)
- There's no particular reason. At that point it's pretty easy to see the solution either way, and Sal apparently didn't feel the need to factor out (n+1).(14 votes)

- why can't we conclude using this test whether the series converges or diverges when the lim of the ratio as n->∞ = 1 ?(14 votes)
- The way the ratio test works is by evaluating the absolute value of the ratio when applied after a very large number of times (tending to infinity), regardless of the initial terms in the series.

If the ratio near infinity is less than 1, then we know for certain that each term is becoming less and less and the series will converge. If the ratio near infinity is greater than 1, then we know that each term will continue to grow, so the series will diverge.

But if the ratio near infinity is equal to 1, then the ratio will not cause the initial terms of the series to either grow or become smaller, so the convergence or divergence of the series cannot be determined by the ratio, additional information about the first terms of the series is needed to decide. This is why when the ratio test give a 1, the results are inconclusive.(19 votes)

- Does this work for only geometric series?(6 votes)
- I think it works on all series, as Sal used it on a series that wasn't a geometric series.(8 votes)

- So I understand the ratio and divergence test, but I don't quite know when to use the tests? Is there something I should look for to signify which test to use?(5 votes)
- Unlike Ratio test, you cannot determine if a series is convergent from the divergent test. Even if the divergent test fails . it does not mean the series is convergent( eg: take the series sigma 1/n).

I would start with the ratio test, because it seems more definitive.(3 votes)

- What is a factorial "!"?(4 votes)
- n! is the product of all positive integers from 1 to n.

3! = 1·2·3 = 6

5! = 1·2·3·4·5 = 120

More here for definition and uses:

https://www.mathsisfun.com/numbers/factorial.html

https://en.wikipedia.org/wiki/Factorial

http://mathworld.wolfram.com/Factorial.html(4 votes)

- If L is equal to 1 why is it inconclusive ? Don't we know that a series which does not converge ,diverge . So if common ratio is approaching 1 then it should diverge as there is no in between .

Also how do we know tht n^10/n! is in geometric progression ,or this test applies to any series other than G.P?

~thankyou(2 votes)- 1 / n diverges, with L = 1; 1 / n^2 converges, also with L = 1. Just with these two examples, we have shown that when L = 1, we cannot be sure of convergence or divergence.

n^10 / n! is definitely not geometric, but the ratio test applies to all series. The geometric series test is just a specific case of the ratio test.(5 votes)

- limit n->inf: (25^n)/(2^n^2).

(25^n+1)/(2^(n+1)^2) * (2^n^2)/(25^n) = (25*25^n)/((2^(n+1)^2) * (2^n^2)/(25^n)

= 25* [ (2^n)/(2*2^n) ]^2 =25*(1/2)^2 = 25/4

My online homework says that this is incorrect. Where did I go wrong?(2 votes) - Hi! At5:36, how does the infinity turn to 0?(3 votes)
- So, the numerator is a massive polynomial, but the largest term is n^10, so it will be n^10 + an^9 +bn^8 + ... + z where a through z are some real numbers. The denominator is n^11 + n^10. You could maybe look at it as n^10 + n^11. the n^10 in the numerator and denominator kinda cancel out, or at least they will be the same number no matter what n is, so we only need to worry about the rest. Nowwhat happens as n gets bigger and bigger? n^11 will always be bigger than an^9 + bn^8 + ... + z. the numerator may start out bigger, but as you head toward infinity, the larger exponent will always make a bigger term, so the denominator will get bigger than the numerator. What does that mean? well, a large denominator makes the fraction get closer and closer to 0. if the denominator keeps getting bigger than the numerator than eventually it will equal 0.

If you'd like to see it on a smaller scale, try (100x^2 + 100x)/(x^3) it starts with the numerator being larger, but eventually the denominator is bigger. Keep plugging in values if you want.

Let me know if that didn't fully help and I can try explaining differently(3 votes)

## Video transcript

- [Voiceover] We already have a lot of experience with the geometric series. For example, if I have the
infinite geometric series starting at N equals K to
infinity of R to the N, which would be R to the K
plus R to the K plus one plus R to the K plus two and keep going on and on and on forever. There are a few things we've
already thought about here. We know what the common ratio is. So the common ratio,
which is the ratio between consecutive terms, is going
to be R to the K plus one. Actually, let me just write
this as R to the N plus one. I don't want to fixate on
this K right over here. So R to the N plus one over R to the N. Well, this is just going to be equal to R. Anything to the N plus one over
that same thing with the N, that's just going to be equal
to R, or R to the first power, and you see that here. When you go from one term to another you are just multiplying, you
are just multiplying by R, and this is all review. If it is not review, I encourage you to watch the videos on geometric series. And what's interesting
about this is we've proven to ourselves, in the videos
about geometric series, that if the common ratio,
if the absolute value of the common ratio is less than one, then the series converges. And if the absolute value of R is greater than or equal to one, then the series diverges. And that makes sense,
we've proven it as well, but it also makes
logical sense that, look, if that absolute value
of R is less than one then each term here is going
to go down by that common, or it's going to be multiplied
by that common ratio, and it's going to decrease
on and on and on and on until it makes sense
that, even though this is an infinite sum, it will
converge to a finite value. Now, with that out of the way for review, let's tackle something a
little bit more interesting. Let's say that we want to look at a, we want to figure out
whether a series like this, so starting at N equals five to infinity of, let's say, N to the tenth power. The numerator is growing quickly. N to the tenth power over N factorial, and factorial we know also
grows very, very quickly, probably it grows much faster than even a high degree polynomial like this, or a high degree term like this, but how do we prove that it converges? We could definitely the diverges test to show that this does not diverge, but how do we prove that
this actually converges? And maybe we can use a little
bit of our intuition here. Well, let's see if we can
come up with a common ratio. So let's see if there's
a common ratio here. So let's take the N plus oneth term, which is going to be N
plus one to the tenth power over N plus one factorial and
divide that by the Nth term, so let's divide that by N to
the tenth over N factorial. Well, dividing by a fraction,
or dividing by anything, is equivalent to multiplying
by it's reciprocal, So let's just multiply by the reciprocal, so times N factorial over N to the tenth. Remember all I'm trying to do
is exactly right over here. See if there is some
type of a common ratio. Well, if we do a little algebra here N plus one factorial, and
factorial algebra is always fun. This is the same thing as N
plus one times N factorial. Times N factorial. You're not use to seeing order of operations with factorials, but this factorial only applies
to this N right over here, and why is that useful? Because this N factorial
cancels with this N factorial, and we're left with N plus
one to the tenth power over N plus one times N to the tenth. Times N to the tenth. So I know what you're thinking. "Hey, wait, look, this isn't
a fixed common ratio here." The ratio between consecutive terms here, when I took the N plus oneth
term divided by the Nth term, it's changing depending on what my N is. This ratio, I guess you
could say, is a function of N so this doesn't seem too useful, but what if I were to say, "O.K., well look, with any of these series "we really care about
the behavior as our N's "get really, really, really large as the limit, as our N's, go to infinity." So what if we were to
look at the behavior here, and if this is approaching some actual values as N approaches infinity, well, it would make conceptual sense that we could kind of think of that as the limit of our common ratio. So let's do that. Let's take the limit as N
approaches infinity of this thing. So remember what we're doing here. This isn't just kind of some voodoo. Conceptually, let me copy and paste this. Conceptually, we're just trying to think about what is the common ratio approach? What is the ratio between
consecutive terms? The N plus one term and the Nth term as N gets really, really,
really, really large? And what we see here is
we have this up here, we don't even have to multiply it out, this is going to be N to the tenth plus a bunch of other stuff. It's going to be a
tenth degree polynomial, and this down here is going to be, actually you can figure it out, this is going to be N to the
11th plus N to the tenth power, so the limit as N approaches infinity, well this denominator is going
to grow faster than that. You could divide the numerator and the denominator by N to the tenth, actually the numerator and the
denominator by N to the 11th, and everything up here
is going to go to zero, and this is going to be one, and so the limit, right over here, this limit is going to be equal to... This limit is going to be equal to zero. As N approaches infinity one way the ratio between consecutive terms approaches zero. So this seems, if we use
the logic from our common ratio up here when we
looked at geometric series, this is clearly not a geometric series, but we would say, "Well, look, as N gets really, really, really
large the ratio between "consecutive terms gets smaller
and smaller and smaller. "Hey, maybe we could
make the same conclusion. "Maybe we could make
the same conclusion that "therefore this series
actually converges." The series actually converges,
so N factorial converges. So can we make this statement? And the answer is yes we can, and what allows us to
do it is the ratio test. Is the ratio test. Let me write that down. The ratio test. And all the ratio test tells us is if we have an infinite series... If we have an infinite series, and I can say from N equals K to infinity, and if we take the limit
as N approaches infinity, of the absolute value
of A to the N plus one over A to the N, and over here I didn't take the absolute value of that limit, but we absolutely could have taken the absolute value of that limit if all of these terms right
over here were positive, so the absolute value would
be the same thing as itself. If this approaches some limit, if this approaches some limit, so once again, this limit is
what is going to be our ratio between consecutive terms
as N gets larger and larger, the ratio tells us if L is less than one, which was the situation here,
L is clearly less than one, then the series converges. Some definitions will say
it's absolutely convergent, it converges absolutely,
the absolute value of the series converges,
but then that also means that the series itself converges. If L is greater than one
then series diverges, Series diverges. and if L is equal to one it's
inconclusive, we don't know. We would have to do some other test to prove whether it converges or diverges. So that's the essence of the ratio test. Find the absolute value of the ratio between consecutive terms, take the limit as N approaches infinity, if that approaches an actual limit, and that limit is less than
one, then the series converges, and it's really based on
the same fundamental idea that we saw with the common
ratio of geometric series.