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25.9 (a) Let 0 < a < 1; Show the series CR=0 converges uniformly I-a.a] to(b) Does the series Cn=" converge uniformly on (-1.1) to Explain....

Question

25.9 (a) Let 0 < a < 1; Show the series CR=0 converges uniformly I-a.a] to(b) Does the series Cn=" converge uniformly on (-1.1) to Explain.

25.9 (a) Let 0 < a < 1; Show the series CR=0 converges uniformly I-a.a] to (b) Does the series Cn=" converge uniformly on (-1.1) to Explain.



Answers

The infinite series $a_{1}+a_{2}+a_{3}+\cdots$ has partial sums given by $S_{n}=n-\frac{1}{n}$. (a) Find $\sum_{k=1}^{10} a_{k}$. (b) Does the infinite series converge? If so, to what value does it converge?

Hello. So here is our given series. Now we can observe this uh here that for N greater than or equal to one um we are going to have that the natural log of N to the 100th power is going to be well less than N To the 0.01. So we have here that the natural log of N to the 100 power divided by N to the 1.1 is going to be less than an end to the 0.1 over N to the 1.1. And this year is going to be equal to one over N to the 1.9 Now we have that the series um the series where we have N going from one to infinity of one over N to the um 1.9 Well this converges right because this is a P series at 1.9 is going to be greater than one. So this converges therefore by the comparison test we have a series which is always going to be less than a series that converges. Therefore our given series must converge as well.

Right, okay. So here we are, given that for sufficiently large and that the natural log of N is going to be less than or equal to end to the A. For and greater than zero. Therefore we have that the natural log of N to the 10th power is going to be less than And to end to the a to the 10th power. Therefore we have here that one over the natural log of end to the 10th power is going to be greater than one over. And to the 10. Hey, And if we just let 10 a. b. equal to p. So let 10 A. B equal to P. Then we have that one over the natural log of N to the 10th power is going to be greater than one over N. To the P. Um And for zero Less than P less than one. You know that Uh series and going from one to infinity of one over end of the P is divergent. Um by the comparison test, therefore our given series one over the natural log of end to the 10th um must be divergent. Yeah.

Want to do is we want to actually um consider this following a series which is the sum of n equal want infinity of negative one raised to the N plus one power a seven. Where a seven is actually going to be one of the square we demand for for um odd values of an and one of the N squared for even values of an and what we wanted to um determine first and foremost is does um this series meet the alternating series tests conditions. So here are the conditions for the alternating series test. Um and so does it meet um these conditions. And so the first thing is okay. So we know um if we check the first condition, the limit as anger goes to infinity of one over the square root of van desire equal zero. And yes it does. And we also have the limit as an ghost of infinity of one over N squared. And that also equals zero. So in it does meet that first condition but now we also need to take a look at the second condition that the next excessive term should be less than or equal to that previous term. And so if I go ahead and write out some of my term, so if I just start here and do um negative one and um plus one a seven. So if in is one we should have a one. If in is to then this is going to be a negative 18 and if in is three um which would be odd, then we actually have one over the square root of three. And if N is four then we also have um one over. And of course if N is four, this would be four squared, which is one that should actually be of two. Right? Yeah. Oh no, this should be a cubed. My so sorry I should have written here, this is going to be a cube down here, that is where I messed up. Um And so but even that if I cube this down here that still goes to zero so that is my fault for doing that. Um And so that's gonna be 184 cube, this is 64 then we have a plus, that would be +51 over the square root up five and so forth. Right? And so if I take a look at any successive terms um I noticed that um on this term if I look at the magnitudes and those terms um won over the screw to three is actually greater than 18 So if I do one divided by that is actually point 577 where this is point 125 So my ace of in term is actually less than my ace of N plus one. And I don't I can't have that right. I have to have my next term has to be less than my previous term. So it does not meet the condition. So this does not meet that the ace of N plus one term has to be less or equal to the case of intern. Okay, so now um does the series converge, So now we want to do does the series converge and the answer is going to be no it does not. And it is because if I actually look at this okay um it is actually made up of um P series. Right? So this is one over into the P where this P is greater than one. Um And so this part does convert but also is made up of one of this group of in or one over into the one half. And in this case P. Is less than one. So this part actually diverges. And so since this series is made up of both parts, it will also um diverge.

This question is asking us to find the some of the 1st 10 numbers in a geometric series given by the partial sums and minus one over end. I know it's asking me for the 1st 10 numbers in the series, because the summation goes from 1 to 10, plugging in 10 for n I'm going to get 10 minus 1/10, or simply nine and 9/10 or 9.9 for my value of A. Then it asks us does the series converge? So what happens if an keeps getting bigger and bigger? Let's find the limit if an approaches infinity of an minus one over and well and is going to go all the way up to infinity and one over end is going to get progressively smaller all the way to zero. This will ultimately lead me up to an infinite amount, which means the series does not converge. In fact, it diverges.


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