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Determine whether or not the series converges. If it converges, find the sum: It is helpful to write out the first few terms of the series...

Question

Determine whether or not the series converges. If it converges, find the sum: It is helpful to write out the first few terms of the series

Determine whether or not the series converges. If it converges, find the sum: It is helpful to write out the first few terms of the series



Answers

Indicate whether the given series converges or diverges. If it converges, find its sum. Hint: It may help you to write out the first few terms of the series. $$\sum_{k=1}^{\infty}\left(\frac{e}{\pi}\right)^{k+1}$$

So her this problem, we need to indicate whether the given series converges or diverges and if the series converges then we have to find its some. Now just given series we have sigma K equals to two to infinite One over K -1 over K -1. Now we have to see whether this serious converges or diverges for that. We're going to expand the series second plug chaos to hear because K starts from two, so 1/2 minus and here it would become one plus. Now we can put Ks three, so we'll get 1/3 Plus. Sorry, this is -1/2 and we're going to repeat like this. And now when we put K as N -1, so it will be one over N -1- Here, it would become one over N -2. And now if you put Ks in so it will be one over N -1 Over N -1. No, here we see that this 1/2 and this negative 1/2 will cancer out like this. And here also this plus this one is plus plus end over one over n minus one will cancel out by this one over n minus one. And finally we can write this series, us negative one plus one over N. So these terms we're going to lift other times we're going to cancel out. Now. If first of all, you should know that how we are able to determine that the series converges or diverges. Now if limit an approach is infinite. A. And If this value is zero then the series converges and if limit and approaches and for night A. N. Is not equal to zero. Then the series diverges. This is the way we determine whether the city's converge or diverge now here A. And we have this much. This one is A. N. So let us find the limit. So limit. An approach is infinite. A. And we have negative one plus one over N. Which will be called, so we can put in as infinity here, so it will be one over infinity, which is zero. So here we are going to lift with negative one, which is not equal to zero. Therefore the series we're going to diverge. So this is the solution for the given problem. I hope you have understood the problem. Thank you.

So here in this problem we need to indicate that whether the given series converges or diverges and if the series converges then we have to find some of the series and the given series is sigma okay equals to one to infinite K. Factorial divide by 100 to the power K. So this is a series we are given with now I want to expand the series and for that we can plug Chaos one, then two and three and so on. So it will be cool when you plug Chaos one, it will be one factorial means one divide by 100 plus. When we put KS two will get two factorial which is one. So to divide by 100 to the power to plus when we put chaos three will get a three victoria which is a six uh street. This would become six divide by 100 to the power three. And the series will goes on like this. Now if you consider in the in term of the series, that is A. N. It will be cool too. And factorial divided by 100 to the power end. And And it plus one when it's time would be that means N plus one this term would be and plus one factorial Divide by 100 to the power n plus one which can also be dinners this N plus one times. And factorial we can write this way also divide by 100 times 100 to the power in. We can write this value in this way also. And if you consider this one this is this is nothing but an item that is A. N. So we can write that N plus one term that is a. N plus one is nothing but it is a cold stew and times N plus one, divide by 100. We can write this way also And if the value of N is greater than 99, then in its plus one, this term will be greater than its previous term that is A. And trump. So we can conclude that the series is actually increasing or other way around. We can also say that the limit an approach is infinite, A. N. Is not equal to zero. And since it is not equal to zero, the series, the series diverges. Therefore, we can conclude that the series diverges and we cannot find it some. So I hope you have understood the problem. Thank you.

So here in this problem we need to indicate that whether the given series converges or diverges and if the series converges then we have to find the some of the cities. Now the given series we have a sigma Okay equals to 1 to infinite. Two over K-us two times. Keep. No I want to expand this series and you can start plugging the value of K. As one then two then three and so on. Now if you put Ks one will get to over three and when you put Ks too so we'll get to over two plus two is four times two is eight. So this would be to over eight plus. Now we put K. S. Three will get to over 15 plus and soon this will goes on like this Now a big and write this to over three can be written us 1 -1/3. One minus 1/3 means to over three. So we can write this way also bless now this one over it to over eight can be written as 1/2 minus 1/4. Like this plus This to over 15 can be written as one over 3 -1/5. And it goes on like this. And if you write the N -11 term it will be So this plus plus plus and it will go on like this up to N -1 term would be one over N -1 plus one over N plus one. Plus the any term would be one over N -1 over and plus two. Now, if you see carefully this series, this 1/3 and 1/3 will cancel out. And we can write this series as This will be equals two, one plus 1/2 -1 over n plus one -1 over n plus two. Which can photo be written us 3/2 minus two. N plus three, divide by N Plus one times and plus two. Now this is basically the entire system now to find that whether this series converges or diverges, we know that limit and approaches infinite A. N. If this value is equal to zero then the series converges. And if the limit an approach is infinite, A. N. is not equal to zero, then the series diverges. Now let us check this limit for this. A. In So we have a limit and approaches infinite A. And we have this much. So this would be 3/2 minus. This is two N plus three, Divide by n plus one Times and Plus two. Now we can also write this limit us 3/2- limit and approaches infinite. This is two, n plus three, divide by. This is N plus one times and blessed to. And if you multiply these two terms will get in square Bliss three n Plus two. Now I'm going to divide with end square in the numerator and the denominator. So this limit will be called to. This is 3/2- limit and approaches and for night. So I want to divide it with N square. So it will be too over in Plus three over end square divide by here it would be one plus. This is three over in Plus two over end. Now, as the end approaches infinite, this total town would be zero. This one will be zero. So finally we'll get the answer to be 3/2 and which is not equal to zero. Therefore the limit is not equal to zero. That means the series we're going to diverge so we can conclude that the series we're going to diverge. So this is the solution for the given problem. I hope you have understood the problem. Thank you

So here in this problem we are given a series which is sigma K equals to one to infinite negative 1/4. Hold to the power negative k minus two. Now we have to indicate whether this series is a diverging series or a converging series and if the series converges then we have to find it some. Now let me first right, it's few terms. So when we plug K as one, so it would be negative one over fall. Hold the power negative three like this plus negative 1/4. Now we can put chaos to her. So it would be holding the power minus four plus negative 1/4. Hold with the power now we can put chaos through here. So it would be negative five and it'll go as like this. Now you can further rewrite this series as this would become a negative for to the power cube. This will also begin right and similarly This would be -4 hold to the power food. We're just taken this denominator in the numerator and this negative power would become the positive power plus here also this is negative four. Hold to the power five and it goes on repeating like this here we see that. And this is a geometric series with the first term of the series to be -4 to the power three And the common ratio are would be equal to it is the ratio of second term divide by the first time. So negative for holding the power four divide by negative four. Hold to the power three which is equal to negative four. And if you see the magnitude of the common ratio, it is four which is greater than one. And since the common ratio is greater than one, so the series will diverge because the series diverges so we cannot find it some. So I hope you have understood the problem. Thank you.


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