Define clearly what it means for sequence {Sn to have lim Sn this case the sequence {3sn} also has limitt +o_+0_ Prove in...

Proof Suppose that $\Sigma a_{n}$ and $\Sigma b_{n}$ are series with positive terms. Prove that if

$\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=0$

and $\Sigma b_{n}$ converges, then $\Sigma a_{n}$ also converges.

So since we know a N is bigger than zero and we need a eunuch or itchy so that in court see is signed a year in exported and AM so this is the pickers we can ride. It's as Sign a n Miner if you go to Sai a n Miners on there and you things Ah, not long chin on serum. We know it's well be course I Kasai chimes A n minus there And it's weird because time since I sense a n and this well's smarter than A because they're co sign, we're always smarter than what? At night we had the Zeon quality. We can use the Iraq comparison serum and scenes. Some mention a em converters. It's imply some insurance sign A n also converges.

Simitian. I end and we can sit Doon under some sort of bono and square and one of his grace convergent bye. Be good to you, Bridget and won. And now we didn't limit on the in. I'm a one of a square. Listen, the cultural immune under an investor infinity And we had an square temps in the end. And it's really good Jesus, given in the question and therefore by the park on Olympic, um, person test, but being here, we can conclude it does. Um, Sam, I am we'll be convergent.

In this question here, where recorded that building it off that paint and invest infinity. You could, you know, if for absolute greater than zero there exists a constant and so stop for all and greater equal to the n on then the absolute AM minus. L must be smaller than absolute. And in this question here were given the I am ik 02 the one other in about three. And we need to show that the limit on this I am here in the he could to the limit off the one of em. Power three. It must include agenda zero and in it to do that. Now, let's try to consider the absolute after I am minus zero, which is a good you're absolutely one of the empire dreaming a zero because you're absolutely one other. And about three we wanted to be smaller than Absalom. So we need to choose that. We need to choose no and way need to choose have camped in here. Uh, well, it could be. It says that presents the one of the end capital Power trees monitor than absolute isn't implies that nah, and about three must be bigger than one absolutely understand, implies that and must be bigger than the one over with three on the Absalom. So therefore we will see for own in that greater than the one over with three on the Absalom isn't implies that the absolutely in minus zero in coach Absolutely one of the n power three. It will be smaller than the absolute on. Doesn't implies that the limit after I end in goes to infinity. Coaching zero. Yeah. Thank you.

Okay, so here are given series is we have a series, we have end going from one to infinity of N. Factorial over N to the end. That gives us a a sub N is equal to N. Factorial over N to the end. Then from the ratio test we then would evaluate the limit as N goes to infinity of a sub N plus one over a sub N. So that gives us the limit as N goes to infinity of n plus one factorial um divided by N plus one to the N plus one and then times and to the N divided by N factorial, which gives us the limit as N goes to infinity of and over N plus one to the end, which um here is going to be equal to well it's gonna be equal to one over the limit as N goes to infinity of one plus one over and to the end. And we know that E is equal to the limit as N goes to infinity of one plus one over N to the end. So therefore this limit here is going to be equal to one over the number E. This is one overeat or one over E is less than one, so one over easy. Right? We know that this one over is less than one, so therefore by the ratio test, um this series converges okay. And then therefore by the end term test for divergence, we know that if we have a series where we have and going from one to infinity of a C event in this series converges, then we know that the limit of the actual terms or limit as N goes to infinity of a seven must be equal to zero. Okay. But the terms um so the terms of the sequence must approach zero. This implies the limit as N approaches infinity of N. Factorial over N. to the end is equal to zero. So again, this implies that the limit as an approaches infinity of and factorial over and to the end, while must be equal to zero. Yes, Yeah.