5

Point) Consider the series Cn On where(~)"n n2 + 4n + 5anIn this problem You must attempt to use the Ratio Test to decide whether the series converges_ComputeO...

Question

Point) Consider the series Cn On where(~)"n n2 + 4n + 5anIn this problem You must attempt to use the Ratio Test to decide whether the series converges_ComputeOn]1 L = lim n+OCEnter the numericab value of the limit L if it converges, INF if it diverges to infinity; MINF if it diverges negative infinity: or DIV if it diverges but not to infinity or negative infinity:Which of the following statements true? The Ratio Test says that the series converges absolutely: B. The Ratio Test says that th

point) Consider the series Cn On where (~)"n n2 + 4n + 5 an In this problem You must attempt to use the Ratio Test to decide whether the series converges_ Compute On]1 L = lim n+OC Enter the numericab value of the limit L if it converges, INF if it diverges to infinity; MINF if it diverges negative infinity: or DIV if it diverges but not to infinity or negative infinity: Which of the following statements true? The Ratio Test says that the series converges absolutely: B. The Ratio Test says that the series diverges_ C. The Ratio Test says that the series converges conditionally: D. The Ratio Test is inconclusive, but the series converges absolutely by another test or tests E. The Ratio Test inconclusive but the series diverges by another test or tests F The Ratio Test inconclusive , but the series converges conditionally by another test or tests_ Enter the letter for your choice here



Answers

Apply the Ratio Test to determine convergence or $d i-$ vergence, or state that the Ratio Test is inconclusive. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1} n}{5^{n}} $$

Hello. So given our even serious we have that our A seven here is going to be equal to three M. Plus 2/5 N cubed plus one. Therefore a sub N plus one is we just replace and with M plus one. So a sub N plus one is going to be equal to well three times and plus one plus two and then divided by five times and plus one cubed plus one. Okay. And then for the ratio test will compute um well we first we want the ratio of a sub N plus one over a sub N. So a sub N plus one over a sub N. That is going to be equal to. Or we get a three M plus 5/5 times N +12 plus one. And then multiplying by the reciprocal of A seven. That's five and Q plus 1/3 M plus two. And that's going to give us a 15 N. to leave 4th plus a. 25 and cubed plus a three N plus five. All divided. Bye. The quantity here, five N cubed plus 15 N squared plus 15 N plus six times three N plus two. Okay, so you multiply this out. Um and combine like terms and then we end up with a 15 N julie. Fourth plus 25 N cubed plus three N plus five. All divided by a 15 and to the fourth plus a 55 N cubed plus a 75 N squared plus 48 N plus 12. Okay, so that is a seven plus one over a sub N. We then want to compute the limit so that we want that row here is going to be equal to the limit as N tends to infinity of the absolute value of a sub N plus one over a sub N. So that is then going to give us the limit as N tends to infinity of what we have up here. We in fact brought an end to the fourth, so we have end to the fourth times 15 um and then plus 25 over. N plus three over and cubed plus five over And to the fourth. And that would then be divided by and to the 4th times 15 plus 55 over. n plus 75 over and squared plus 48 over and cubed plus 12 over. And to the forest. Thanks. Okay. Um and we then we take the limit here well um and to the fourth over, end of the fourth, right? That just becomes one all of this becomes 0000, this all goes to zero. Um so we just get 15 plus a bunch of Zeros Times one. Um and then 15 plus of buses, Zeros Times one. This just becomes 15/15, which is one. So here we have that row is going to be equal to one. Um and since we have equal to one that tells us that the ratio test then is inconclusive right? If we're greater than one or less than one, we can make conclusions if we have convergence, but if we're equal to one, well then we say the ratio test in this case is going to be in conclusive and we know nothing. All right. Take care. Two.

Hello. So from a given series we have any going from two to infinity of one over the natural log event. We're gonna have a seven here. Is that we equal to one over the natural log of n. And then for the ratio test we want to compute well the we want the absolutely limit as N goes to infinity of the absolute value of a sub N plus one over a sub N. So the absolute value of a sub and plus one over a sub N is going to be equal to while we get one over the natural log of M plus one times Ellen event, which is going to be equal to the natural log of N over the natural log of n plus one. So then taking the limit um we get the limit here as N goes to infinity of well the absolute value of of a seven plus one over a sub N gives us the limit as N goes to infinity of Ellen event over Ellen of m plus one gives us the limit as N goes to infinity of one over and over one over and Plus one using Lopa towels rule because we plug in um, infinity here we get infinity we get intimate form so we can use local towers role, take the derivative of the top over the deliver of the bottom and then we end up here with the limit as N goes to infinity of just well, N plus one over N. Um, which then gives us the limit as N goes to infinity of just one plus one over N. So then taking the limit here as N goes to infinity one over end that goes to zero. So therefore the limit here is going to be equal to one. So, since we have um, the limit as N goes to infinity of hmm plus one over a sub N. Since that equal is equal to one. In this case, if it's greater than one or less than one, we can conclude if we're um convergent or divergent, but here we're equal to one. Therefore, the ratio test is in conclusive. All right, take care.

I need to do the immigration asked, man, we need to compute the limit and just infinity under my endless won the money. But I am And we go to the limit and just to infinity. And here I am, this one We're ego to empress one em's fiber. And this one dividing by He had a and we could teach you And five. And just when we have the element of mystery And now we divide by I am means then we need to move my better reciprocal will be there to em Bliss Three hymns and then off Aimless one The finding by the end times Fine about and yeah, we noticed that there five about and cancer with spam on the top here. And then we have man would, uh, in Cochin limit and just to infinity. And we have two groups near the first group will be and plus one times that to embarrass three You and this five times dame we haven't under limit. And just to infinity. And then we have dessert and enough amber sk one You ready, man of endless Do so Notice that on the first limit here we have the sermon degree on the salmon in the bottom. That's a maximum decree. Therefore, this Rimini coach Rhoda covers, and next to the maximum guy, would you beat you? And isn't it also true here, So from bandit, you are with you. And here for this one. We get, uh, for this limit here. Uh, also, we forgot to bring the three fiving our side. So we have a five year five here and this limit here s Angus to infinity. Gonna oneness. Well, so therefore again defy you mentally and fight credit. And one therefore this seriously and will be on divergent buying a limit ratio task here.

Hello. So here our ace event is going to be an over and squared plus one. And then we have that a sub N plus one over a sub N. Well that is going to be equal to a seven plus one. Is N plus one over and plus one squared plus one. And then we multiplied. Was divided by race event multiple reciprocal here. So we're multiplying by N squared plus one over N. So therefore this is going to be equal to um and n cubed plus and squared Plus N Plus one. All divided by while N squared plus two N plus two times and gives us an N cubed plus two and squared plus two. N. Okay, so there's a seven plus one over a seven. Now we want that row here is going to be equal to the limit as N tends to infinity of the absolute value of a sub N plus one over a sub end. So that is then going to give us the limit as n tends to infinity of well we get nQ plus and squared plus and plus one over N cubed plus two N squared plus two N. Um Which we can factor this as an n cubed times one plus one over N plus one over n squared plus one over and cubed and then divided by an n cubed times one plus one over N plus or plus two over end the way that you matter. But plus um yeah, just give us a hopes uh and cubed times one plus two over and plus two over and squared and then taking the limit here as um and tends to infinity. What do we get? Well we get this is just going to be equal to one plus a bunch of zeros times one. This limit here is going to be equal to one. So we have that role then is equal to one. And since roe is equal to one that tells us well, doesn't touch anything, it tells us that the ratio test in this case is going to be inconclusive. Okay. Yeah. Yeah.


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