Question
B. Use Comparison Test to determine the convergence or divergence of the seriesnvn +1
b. Use Comparison Test to determine the convergence or divergence of the series nvn +1
Answers
Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{1}{n !} $$
Hello. So here we have. The A. Seven is going to be equal to E to the negative end or each negative end. That's one over E to the end. We can apply the root test here and take the limit as N tends to infinity of the N. Fruit of the absolute value of a servant. So that's gonna be the limit. As N goes to infinity of the entire route of the absolute value of one over E. To the end is just going to give us one over easy. The limit as N goes to infinity of just one over E. Well, there's no end anymore, so therefore this is equal to one over E and one over E is less than one. So therefore by the root test that tells us that are given series is going to converge.
Hello. So given series um and 02 infinity of E to the native and squared can be written as the series one over E to the end squared. And since we have that E to the end squared is greater than or equal to eat at the end for all and greater than or greater than zero. Um We get that one over E to the end squared is less than or equal 21 over E to the end. So um well looking at one over E to the end, that's going to be a geometric series. Right? One over E to the end, one over E to the end is a convergent geometric series with common ratio here um are going to be equal to one over E which is less than one. So this is a convergent geometric series. Therefore our given series is also going to converge by the comparison test.
Here's to find either convergence or divergence of serious. And so one of the tests were going to be using here. Let's just go ahead and take the limit as an approaches infinity. Let's see what we get here. And so this If I multiply The one over and to top and bottom. So I get the limit as N approaches infinity of 1/1 plus one over N splitting that up against the limit has and approaches infinity to the top on. And as N approaches infinity of one, this alignment has approaches infinity of whatever. And so that would be 1/1 plus zero chuckles one. And any time a limit goes to as it goes to infinity equals anything other than cereal that we know that per the divergence test, serious type purchase. That's your answer there.
Hello. So here we have a series and going from one to infinity of one over N squared plus one. Now since we have that N squared plus one is greater than or equal to and squared for all and we have that one over and squared plus one is going to be less than or equal to one over and squared. So we get that the sum and going from one to infinity of one over N squared plus one is going to be less than or equal to some and going from one to infinity of one over N squared but we know that one over and squared is a P series right? That is a P series where P is equal to two which is greater than one um which is going to converge. Therefore by the comparison test, our given series is also going to be conversion.