All right
Here are a few more examples using the direct
Comparison test or the limit comparison test to show whether a particular series will converge or diverge
[in] this case notice we have [one] plus sine of n over 10 to the n
The trick a lot of times if you have sine, or cosine is
Recall that sine is always between positive 1 and Negative 1
so that means if we add 1 to the Middle well 1 plus sine of n is always between positive 2 and
0 and
that means that 1 plus
Sine of n over 10 to the n. That's going to be less than [or] equal to
2 over 10 raised [to] the N for all n
for all n [greater] than or equal to 1 so
[the] idea in this case is
Well 2 over 10 to the n. Do we know anything about that particular series and
Got another example down below so let me
squeeze this one in here
well if you think about
Again the series 2 over 10 to the N
You can always factor the 2 out of this
Notice the 2 doesn't have an [n] on it so I can [just] pull that out front
And then I'm left with 1 over 10 to the N
hopefully, this is starting to look a little more familiar and
the trick is you can write 1 as
1 to the N, because 1 to any power is still 1 so I can't a can in on top of that
But now I can rewrite [this] one one more time
So might choose out front. I've got [1/10] raised to the N, and
this is now a geometric series and
This is going to converge because my number 1/10 is smaller than 1
So this is a convergent geometric series. We've showed that 1 plus sine of N
So don't look at my next example. We showed that 1 plus sine of n over 10 to the N
Is smaller than that well since the geometric series converges and this one is smaller this series will [also]
Converge as well
So there's an example of the direct comparison test
On my next series here my next example
I'm going to use
From [N]. Equals 2 to infinity of square root of n over n. And again the thing I'm going to compare this one to I
Take the highest power on top the highest power on the bottom
well this is n to the 1/2 over n to the first this is equivalent to n to the 1/2 and
again if I think about this as being a series and put the summation out in front of the 1 over [n] to the 1/2
that is [a]
Divergent P series again recall that this power has to be strictly greater than 1
Well, if I look at the limit as n goes to infinity of the a sub n
That's square root of n. Over [n]. [minus] 1 if I divide that if I make this stuff my b sub N
if I divide all of that that's going to be equivalent to multiplying by
into the [1/2] over [1] and
This is going to give me the limit as n [goes] to infinity
This is Square root of N times square root of N. That will give me [N]
Nothing's going on in the denominator
again as n goes to infinity
Infinity the highest power on top is the same as the highest power on the bottom I can just take the ratio of their
coefficients and get this limit to be 1
So again in this case. It says that both of these series will do the same thing
since I compared it to a
Divergent p
Series that again means that this original series that we started with must also be divergent as well
Okay
Let's look at one more example here
So in this case we have n. Factorial over n, raised to the N, and
You know sometimes it's good [to] get a feel [for] these
So maybe plug in a couple terms obviously and recall that you know a factorial for example 5 factorial
That's just 5 times 4 times [3] times 2 Times 1
So we'll get 1 over 1 that'll be our first term I'll get 2 times 1 I plug in a 2 I'll get 2 times
2 if I plug in a 3 I'll get 3 times 2 times 1 over 3 times 3 Times 3
ETC 4 times 3 times 2 times 1 over 4 Times 4 Times 4 [Times] 4
So intuitively to me, it seems like each
Term is probably getting smaller because notice the first number is
Always going to be equal to 1 but then you're multiplying it by something smaller than 1
multiplying it by something smaller than 1
Multiplying it by [something] smaller than 1 so it's going to keep making this number shrink
So certainly the terms are getting smaller, but again that doesn't justify whether it converges or diverges?
But let's write out in Factorial
So in factorial over n to the n. If we write it out Generic. I'll have n times n minus [1]
Times N minus 2
All the way down to 3 times 2 times 1 and obviously this is assuming that n. Is relatively large
So then I'll have N
Times N
Times N times N times [N]
and
Let's think about this one for a second
So we're getting one here at the beginning
This is always something less than or equal to one
This is always something less than or equal to one ETC all the way down the line
and
that
[means] in that case I can claim [that] n. Factorial over n to the n is [less] than or equal to
Well the first term is 1 this is something smaller than 1 the next 1 something smaller than 1 the second one isn't something smaller
Than 1 so everything is going to be smaller than 1
Including the last two terms, but I'm actually going to keep those I
Have 2 over N. Times 1 over N
And again I can conclude this because everything in the middle that I'm kind of leaving out is a number smaller [than] 1
[ok] and
2 over N. Times 1 over [N]. That's equivalent to 2 over N. Squared
But again if we think about this now as being a p-series
2 over N. Squared is equivalent to
We can pull the [n] out [front] and have 1 over n. Squared as n goes to 1 to infinity
And again, this is an example of a convergent [P] series
Okay, so I've now shown that my original
series that I started off with
For all n greater than or equal to one this is less than or [equal] to two over n
Squared well since my p series converges that implies [that] the [original] series is convergent [as]
well
Okay, so this one's certainly not as straightforward. You know you've got some factorials and these this power of N
You can into this power of n. So in this case
You know if you don't have a feel for it write out a few terms and see what's happening and again
This is one where [you] have to think [about] things. You know it's not completely mechanical
You know you really have to argue and convince yourself that this inequality that we've done here at the bottom is in fact correct
but once you have that you're off and running and
you've got a convergent [P] series and
Like I said this last one. I think is a little tricky take a look at it
You know write out a few more terms. If it is baffling a little bit to you and
See if you can't convince yourself of this last part