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(a) State and prove the theorem giving the Weierstrass M-test. You may use the Cauchy convergence criterion without proof:[10](b) Consider the functional series Li ...

Question

(a) State and prove the theorem giving the Weierstrass M-test. You may use the Cauchy convergence criterion without proof:[10](b) Consider the functional series Li (c € (1,0)) J=] Given the information that the functional series (*) is pointwise convergent, show that it is uniformly convergent on [@, 0o) for each (ii) Show that the formally differentiated series (obtained by differentiating term by term) is uniformly convergent on [a,0) for each You may use without proof the inequality j Fa /

(a) State and prove the theorem giving the Weierstrass M-test. You may use the Cauchy convergence criterion without proof: [10] (b) Consider the functional series Li (c € (1,0)) J=] Given the information that the functional series (*) is pointwise convergent, show that it is uniformly convergent on [@, 0o) for each (ii) Show that the formally differentiated series (obtained by differentiating term by term) is uniformly convergent on [a,0) for each You may use without proof the inequality j Fa / log j < where exp(1) ea (iii) Use the results of and to show that the function ((z) Lj [15] (c € (1,00)) is continuously differentiable-



Answers

Use the integral test to determine if the series converges or diverges

In this problem were given that a sequence essen is such that The magnitude of difference between the end plus one term and the entire term is always less than do you raise to the power of negative and so we can consider the maximum difference rather the magnitude of the maximum difference. Mhm. Being equal to s to the Brother, 2 to the power negative end. Now a cautious sequence is defined as a sequence in which the subject subsequent terms become arbitrarily close to each other. So If a seven is a sequence than the magnitude of a plus and And a seven plus 1 minus a seven, mhm is equal to zero as N approaches infinity. So they become arbitrarily close to each other. Now to prove that this sequence is a cautious sequence. We can dig the limit as N tends to infinity on both sides. Yeah. Mhm Okay, okay, Okay. Mhm. What people saw limit uh and tends to infinity of due to the negative end and we can solve for the limit. What the fuck? Mhm. Okay. And it's actually a pretty easy limit to solve. So that's the limit as intends to infinity of two to the end. Now, as N approaches infinity, due to the end also approaches infinity, so and is in the in the exponent and as and increases due to the end blows up to infinity and that means that one over to the end. Yeah, Ghost zero. So our limit evaluates 20. Yeah. So the limit as n tends to infinity As 7-plus 1 minus S seven Rather their magnitude equal zero and therefore as intends to infinity the subsequent terms right, subsequent terms Of the sequence, S seven become arbitrarily close, become are bit rarely close, and therefore the given sequence is a cause she sequence. And since cautious sequences always can converge so cause she sequences in words follows that the sequence S N. Don't forget that also. Yeah. Okay. And bridges and there is our solution to this broad.

The given series is sigma and as he could want infinity minus one power in castle one by an So this is your sees in case the series convergence. So let's assume it is june the series converges. So am I converges, What do you mean by the way in? He is and the democracies which is minus one. Barring goes off one man. And what do you mean by sigma and converges the partial sum. The limit of the partial sum. The sequence of partial ISMs converges to limit, intending doing three D S N is finite. C. M no recall that if a sin is some condoms something first interns then S. N plus one minus sn S A M. The end of the dome some two first and place condoms minus the sum to first and dance will give you another time. Now suppose if limit intending to infinitely ascent is finite, then limit, entertaining to infinitely sent us one will also be finite. Samuel. Now, when you subtract then limit, intending to infinitely S. N. Plus one minus S. N. Should be l minus L. Zero. So that implies limit and tending to infinity you should be zero. So what is the conclusion F sigma A. N converges then limit and tending to infinity A. N. Must be zero. The congress may not be too it's F limit and then into infinity A. And zero sigma N. Need not converge. There's may or may not and watch but F limit and tending to infinity A. N. Is not equal to zero. Then sigma A. In diverges then sing mine diverges. So the contract positive of this statement. Control positive. What is what is contra positive? If B then Q is a statement then it's contra positive is if not cure then not B. So this is the contra possible. So this statement is true then it's contra positive is also true. So what is the contra positive? If suppose this limit is not equal to zero, then this do not converge. Do not converge means diverge. Why I need this now. What is our IAN or IAN is minus one power in cause off one by an let's take limit and and infinitely so when you take the limit its limit and turning to infinity minus one power one man into course off one by infinity, which is zero because zero is one. But what about this? Can it be zero limited entering doing three D minus, empowering. Never. It oscillates its oscillates with two values minus or 91 so it can never be zero. So that means to limit intending to infinity A. N is not equal to zero. So hour series diverges. This is the conclusion.

In this problem we have to find interval of convergence for the given in finite series. So first thing that we can note here, this whole term represents any term of the series, let us say anyth term is represented by B N. So bien is six X. Raised to the power and divided by fifth root of and therefore and plus one term will be we can we have to replace and by end plus one. So this is going to be six times X raised to the power and plus one divided by fifth root of and plus one. No, we will perform raise your taste here. So from ratio taste what raise your taste says? We have to find limit off absolute value of be in divided by B N plus one at an approaches to infinity. If this value is less than one then the city's converse's. And at the end we would have to check the value at the end points. So first we will calculate limit of the function here and then we will calculate the interval. So what is going to be limited? So limit as an approaches to infinity models of being we will put its value six x rays to the power and divided by fifth root of and and what is B N plus one since it is in denominator. So we will reverse the a numerator and denominator term of be off and plus one. So it will be fifth the root of and plus one divided by six times X raised to the power and plus one. Now a few terms will get canceled here and we'll get canceled. Now we can write X raised to the power and plus one X raised to the power and plus one as X in X raised to the power end into X raised to the power one. So simply X. Now this X raised to the power and will get cancelled with the numerator term here. So the left terms will be limit of and approaches to infinity. The numerator terms will be fifth root of n plus one divided by fifth root of and into once upon X. So limit of and approaches to infinity. We can combine these two terms, fifth, the root of endless one upon and into one by X. We can take an comin from the numerator. So limit of this function will be as an approaches to infinity. Fifth wrote off and in 21 plus one by N divided by N into one upon X. Now as here the animal gets canceled now as the end approaches to infinity so this term will approach to zero. Therefore limit value of this entire function will be one therefore, if you perform limit here. So this function reduces to models of one upon X. Now for the conversions we know that the value of the limits should be less than one. Therefore model Fund by X should be less than one. Now from here we get to solutions first solution is minus one by X minus one way X is greater than one and second solution is one by X is less than one, so if minus one where X is greater than one, then X will be less than minus one. And from here we will get X is greater than one. Therefore the interval of convergence is going to be X belongs to minus infinity to minus one union, one to infinity. But we have to check the endpoints here at minus one and one. So accordingly we will modify this result. No, if we dig if exit the calls to one so what will be our cities? So series was energy equals to eight to infinity six and two x rays to the power and divided by fifth root of And so if you put the value of x rays to the power in here. So if you put value of X equals to one here. So this will reduce to sigma. Energy costs 218 to infinity six and 21 raised to the power N divided by fifth root of. And now we know that power of power any number to the power of one will give one. Therefore this will be shake him off and equals to eight to infinity six divided by fifth root of and no we can use we can use divergence taste here. So what divergence taste? It says if you perform divergence test and we take the limit. If we take the limit of the any term of the function here. Six divided by fifth root of n. As N approaches to infinity. So we see that this is a close to zero. It means which is less than one. So if it is, if the limit of the function at any and approaches to infinity is less than one, this implies that the function is conversion. Therefore X equals to one is going to be convergent. So we will use equality at X equals to one. So our so our interval will modify too. X belongs to one to infinity. Now we will check the scene for X equals two minus one. So if exes equals two minus one, so at X equals two minus one. Our cities will modify us. What was our series? It was sigma from any close to eight to infinity. Six x rays to the power and upon fifth root of and that is equals two, six and two minus one raised to the power and divided by fifth root of and and limit as N approaches to infinity. Now if we now if we go for divergence taste then limit as an approaches to infinity of the term that is minus one raised to the power N into six, divided by fifth root of. And we observe here that here the term present is minus one raised to the power end, which is an oscillating limit, which will give an oscillating limit. This implies that limit does not exist. Limit does not exist if limited does not exist. This implies that a taxi cost to minus one. The cities will not be convergent. Therefore that there is the engine of the range of convergence will be X belongs to minus infinity to minus one. There will be an open bracket at taxes equals two minus one union, one to infinity, and one will be included in our solution. Therefore, there will be closed interpreter, so this is going to be our final answer.

Hello guys. So we're gonna answer the question about the equation comes as follows and it goes on to the community two and over and square this one. So we will apply the costume modification rules which is taking out the tour right here bringing it out of the equation times it's gonna be you're gonna look something like this two times and you want to infinitely and over And Square Plus one. Now we're going to apply the limit the series limit compression test, which states that if the limit of N equals infinity uh over N over build A. M equals L. Where zero is less than A. Of N. And B. F. N. And zero is less than L. Or less infinite end less than infinity than the F. N. And F. N. Either both coverage or body virg. So our F. N. It's gonna be Oh yes. And we're gonna be and over and square to us one. Let's go over this one and our B. F. N. It's gonna be one over it. So taking the limit of both. Both of those, we should get we should get a limit. We're going to do a limit of incident for me And over and squared plus one over one over end whenever I'm sorry, they're going to divide fractions True. It's gonna look like this phantoms and over and square plus one times one. If you want to make it, you know, look cleaner, which we have to he's going to do something like this and square over and square plus one because One times x squared plus one is 1 is and corpus one. And so okay, let me and to the infinity and square over And Square Plus one. Number two divided by the highest denominator which is putting a second here and we'll use your side Which is gonna be one over one less one over and square which is the same as square. Where I? M squared plus one. We divide by N square which this is this should be our final answer. This is what our answer should look like. So to get there we'll do and square Over and Square Plus one. You ready back and square if you look like something like this and square over and square. Uh huh. And square over and squared. This one over and squared Becomes 1/1 plus oops those this becomes one over one plus one over and squared. We're gonna take the limit of this too. So let me know and then maybe 104 one small over and squared. And so we will divide both limits. It's gonna be like in front of like this from space we're gonna take the limit of and to the infinity one over let me know. And to infinity one Gloves, one over and squared. Taking the limit of both of those. The limit of one. We know the limit of one is one And the limit of this is also one because taking the limit of one over N square is zero and in the middle of one is one, so 1 to 0 is one. We'll give you one of the 15-1. And if you go back to the limit comparison test criteria, we see that the series diverges. Yeah, serious. These virgins, because we went back to the criteria, he says Where zero is less than a over MBAN and zero is less than L, unless infinity than B, FN and FM. Mhm. You know, both coverages are the body verges. In this case the boat's divergence. Thank you so much.


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