5

Na) Let Tn be an increasing sequence of real numbers that converges to I. Let $ = {Tn In =1,2,3, Show that € = sup S....

Question

Na) Let Tn be an increasing sequence of real numbers that converges to I. Let $ = {Tn In =1,2,3, Show that € = sup S.

Na) Let Tn be an increasing sequence of real numbers that converges to I. Let $ = {Tn In =1,2,3, Show that € = sup S.



Answers

The sequence $s_{n}$ is increasing, the sequence $t_{n}$ converges, and $s_{n} \leq t_{n}$ for all $n .$ Show that $s_{n}$ converges.

So we have a series S a van and we're told that this series is increasing and also we have another series T. Event and we know that this other series is always going to be greater than a seven. So the T. F. N. Is going to be created and recalled to s event. And also we know that the event converges. So having that we need to prove Ah that S. seven also converges well in order to do that, let's look at for example, initial value of is not the first value. Let's just say that's some number right? And because a servant is increasing then the limit as N. Goes to infinity of S. F. N. Is going to be equal to well because it's increasing then it's going to be greater than it is not. But also it's going to be smaller. Then um T. Infinity taking to the limit of T. And since t converges, we know that this is going to be a constant, it's gonna be a number. And so is this, So I have a limit is going to be end up ending up between two um constant numbers to actual values because it's increasing all the way. It's kind of also um converged value. So therefore S. F. N. Also converges

Hello. Children's welcome back. And this question we are given let A N. We are non increasing sequence. Okay. Either it can be decreasing or contract. So it is not increasing sequence of positive real numbers. Then submission of A. N. Where N. Is greater than one. The city's converges. If Submission and greater than zero to to the bar N. And a off to to the parent converges. Okay, we have to prove this. Now this is a serum which is known as Kashi condensation tour. Um Okay so the name of this forum is cause she condensation test. Okay, It is because she condensation test. Okay, so we have to prove this corsican transition test. So for that first of all I have a lead essen is a sequence as in is a sequence of partial sums. Okay, so as soon as a sequence of partial sums off submission A.N. N is greater than one. Okay, now stay in. Is the sequence it is a victim. It is a sequence of passion sums. Oh The other series which is true to the parents and 8 to the bar anywhere and as great as n equals to zero. Now we have to know that S. 17 are increasing sequence because it is increasing. Okay, it is given so a sin and then us increasing sequence. Okay, because it is given now and we need to it's enough to show that they sent For any greater than one is bounded which implies TN will be bounded. Okay, so if we show this then it is uh this uh result will be proved. Okay so first of all we let that S. N. Is bounded. Okay, so we have bled S. N. Where N is greater than equals to one is bounded. So we have already assumed this. Now. We have to show we have to show that CN for any greater german will be bounded. Okay, we have to show this. No uh if I take s to the department equals to even plus a two up to a two part N. Okay, so it becomes even plus a two. So I split a three plus a four and take the rest part aside If I plus a six plus 87 plus eight. Okay And I have taken rest apart so it becomes a two and minus one plus up to eight to depart. And right now we know that this part will be greater than what even plus a do Plus two times of a phone because we know that this is an increasing this is an increasing sequence. So obviously uh sorry we know that this is an decreasing sequence. Non increasing sequence. Okay, I have written wrong here, it is non increasing right? It has given no one increasing so uh you know that since it is not increasing a three will be either greater than a four or equal to a four. So okay, so even plus a two and if I replace a three with a four, so this will become less. So plus to a four plus four times of eight And up to the part, end -1 times of 8 to the bar. And okay, so this will be lesser now moving ahead if I if I did yes to the part and this will be greater than even plus a two plus +24 and up to depart and minus one A two to depart. And right now if I just uh put it like this half of even how whole Even plus a two. So he plans to I do next to it Plus up to the bar and and why I've done this because there was a power and -1 so I have bring the power and okay now we can see that as to the bar and this will be greater than what half of a even and this will become RtN right? If you see this is nothing but the department and a 22 department. Right, so this has become half of PN Now a sense S. S off to the department is greater than this time so we can see that pain is also about it. The end is also bounded. So we have proved the given result. Okay, now uh so from the given results we can say that submission of a N is convergent is converging. It's and only if Submission and greater than 0 to depart en and a of to the party s conversion. Okay, so we can through this result we have proved this result. Now using this result we have to discuss the convergence and divergence of two cities. Okay so the first series given a submission of one upon came to the party. Okay where we are given K is arranging from one to infinity. Okay so we have let let's say am will be one to the past care to the part and okay right so Now we have to find 22 different aid to the department. Right so I will put to do the bar end and at the place of and I'll put to to the parent. Okay so here what I came to the path so to the bar and right so now if I just see it if I just solve this equation you know this is this will always be greater than this stump. Okay so okay to the part to when will always be greater than to the part and so you can say this term this time will tend to infer tonight As an especially this town will tend to zero as intended to us and tends to infinite. How is strong? We came to zero because the denominator is bigger and the numerator is smaller so as you increase and the denominator becomes bigger and bigger. Okay so this whole town will tend to zero as intends to infinite. So we can say that some nation to depart en eight to depart. End converges Converges where end is greater than zero which implies that submission of a and very converged. Okay uh the condition that a man should be uh non increasing that should be always true and we can see that this is a non injury increasing sequence. Okay so we have applied causing condensation test. This is a non increasing sequence right now. The second time which we need to prove is one divided by. Okay, Ellen care to the party. Okay, this is given town where K is ranging from one to infinity. Now we have to solve this uh using condensation test. Okay so now what I'll do is first of all I'll take a N. To be one divided by and Ellen and to the party. Right so this will be our in here. I think I've been some mistake um Yeah here the town will be something like this. Okay I have replaced I have to replace cable to the part and it will be too to the party and to the path P and here the town will be Okay so this is that um no well Go ahead to the 2nd part. This will be our N. So now I know that this is a non increasing sequence. Clearly this is a non increasing sequence. Thank you. Clearly this is a no increasing sequence so we have to check the convergence of submission of to do the part and a to do department. Okay so they will see it so put into the department at the place of end and Ellen off to depart end to the party and here we have two department extra. So this stump got canceled from this stuff and what we are left with is one divided by Ellen to to the baron and will come here and a link to the path we okay so if you see this it becomes one divided by and to the party and Ellen too to the party now if you check this has become something like something which was derived in the first equation. Okay and to the party right so if you check uh if I if I solve first part again okay so there will be uh the explanation will be different now here so explanation will be something like this. This town will cancel out this. So to to the park N A two part end will become one upon through the party. So this always converges. Okay this always converges so we can say some mention of a and will converge. Now if you check when a bone and to the party and into the departed this will converge, this will converge. Yes They will be greater than one. Okay and it will diverge four P. between zero and 1. Okay so since uh to to depart N. A. To depart and is converging for P two P greater than one. So we can say that submission of A. N will converge. Four be greater than one. We cannot say about the divergence of a submission of AM, because caution condensation test only tells about the convergence. Okay, so a submission of AM will converge more pig greater than one. Okay, so that's it. Thank you.

In this question. We need to show that for any real number X. The sequence. One plus ex upon and to the power and converges to here. To the power X. Let's see how to solve this question. Let us consider that. A And it's a sequence and if is a function touch of that if and if equals two Ian four And better than equals to one. No, if LTD K tends to infinity F K If he calls to L then limit and tends to infinity and will also be equal to. And let's say This is equation one. No, if we define F. Okay, if he calls to one plus ex upon K to the power K then we recognize that limit K tends to infinity F K has the indeterminant form I want to be power infinity since K tends to infinity. Therefore to solve this problem, apply the L hospital rule, apply the L hospital rule so we can write limit kate has to infinity log F G. If he calls to limit K tends to infinity. Okay log one plus ex upon. Okay we have taken the liberalism of both the sides. So the sylvie calls to limit, he tends to infinity log one plus x upon k upon bone up on K. No differentiate the numerator and denominator separately. So we get limit Kia tends to infinity one upon one plus x upon key into minus X upon case, choir upon minus one upon He's playing on further solving we get limit K tends to infinity Ex upon one plus Ex upon key hands This will be equals two limit K tends to infinity Ex upon limit okay, tends to infinity one plus ex upon key. Now substitute the limits. So we get limit K tends to infinity log Yeah, F K. If he calls to ex upon one or we can say it will be equals two X. Since limit kate. And for infinity love have physicals two X. Therefore we can right law limit care tends to infinity FK equals two X. And now take the anti lock of both of your sides. So we get limit care tends to infinity F K. If he calls to it was the power X. Now trauma question one equation one. We can write limit and tends to infinity A and he recalls to it would be power X. Mhm. Therefore it can be concluded that limit and tends to infinity one plus ex upon end to the power and if he calls to he to the power X that's the sequence E. N. It recalls to one plus ex upon and to the power and converges two. It was the power X. So this is the final answer for this problem. I hope you're not for the solution. Thank you

All right. So if this works, you know that the absolute value but ace of added plus one divided by a servant and is equal to r to the power off plus one multiplied by an editorial divided by n plus one factorial multiplies by r to the power of that. So this is just equal to the absolute value, are divided by adding plus one if we set, this is equal to s taking the limit As adding approaches infinity of SP end up with zero multiplied by the absolute value off our which is equal to zero, which is less than one. Therefore, this Siri's conversions by the ratio test for old values.


Similar Solved Questions

5 answers
6147For Ihe (eqion bounded byye8-X' Ihe *-axis 1y-4ku determlna #hich ol the Iollo *ing preater: the volume of the sold generaled when ihe regen revohed aboumtthe LatIS aboui the Y-axis
6147 For Ihe (eqion bounded byye8-X' Ihe *-axis 1y-4ku determlna #hich ol the Iollo *ing preater: the volume of the sold generaled when ihe regen revohed aboumtthe LatIS aboui the Y-axis...
5 answers
Determine whether the given informatlon results one; IWQ Or Ihat result; Ho trlangles Solve an; Triangles602. 4 =4 and b = 5.#of triangles:Solution:Solution:31. B = 209 b = 4 and € = 6.#of triangles:Solution:Solution:_
Determine whether the given informatlon results one; IWQ Or Ihat result; Ho trlangles Solve an; Triangles 602. 4 =4 and b = 5. #of triangles: Solution: Solution: 31. B = 209 b = 4 and € = 6. #of triangles: Solution: Solution:_...
4 answers
15.OH1 pointsPrevious Answers SCalcCC4 8.2.039.Express the number as a ratio of integers_ 7.39T0 XNeed Help?Read ItIalktol IutorSubmit AnswerPractice Another Version
15. OH1 points Previous Answers SCalcCC4 8.2.039. Express the number as a ratio of integers_ 7.39T0 X Need Help? Read It Ialktol Iutor Submit Answer Practice Another Version...
5 answers
Exercise 6Is the set31xe B}subspace of R3?ExerciseIs the setsin(x) =0}subspace of R2?Exercise 8Compute19) ( and (1)
Exercise 6 Is the set 31 xe B} subspace of R3? Exercise Is the set sin(x) =0} subspace of R2? Exercise 8 Compute 1 9) ( and ( 1)...
5 answers
To find a relationship belween current and reslstance, what will be the deperdlent variable?CurrentPotential DifferenceResistanceSubmit Answcr submnissions remaining
To find a relationship belween current and reslstance, what will be the deperdlent variable? Current Potential Difference Resistance Submit Answcr submnissions remaining...
5 answers
A) 1 kmh to mlh ?
a) 1 kmh to mlh ?...
5 answers
POINTSLARCALCETZ 14.5.013.Find the area of the surface given by 2 = f(x, Y) that lies above the region R.f{x, Y) = Vx2 + y2 , R = {(x, Y): 0 < flx,Y) < 6}
POINTS LARCALCETZ 14.5.013. Find the area of the surface given by 2 = f(x, Y) that lies above the region R. f{x, Y) = Vx2 + y2 , R = {(x, Y): 0 < flx,Y) < 6}...
5 answers
10 pta. Using tlwee Laplacv Trauslorm, eXprows tlce slution of th" given initial value probkqt in terms of # convolution integral"" + %v "4 cuslat): "() =1,( =0
10 pta. Using tlwee Laplacv Trauslorm, eXprows tlce slution of th" given initial value probkqt in terms of # convolution integral "" + %v "4 cuslat): "() =1,( =0...
5 answers
PeintsBonks ( ptianal ) Fmd Pos tive h4m6ec$ Lsz Pradact I0 I$ 54 Ae #t +ae SLt of 2 times Fhe frst ond 3 tmt +2 Se Cond I$ Iimimun
peints Bonks ( ptianal ) Fmd Pos tive h4m6ec$ Lsz Pradact I0 I$ 54 Ae #t +ae SLt of 2 times Fhe frst ond 3 tmt +2 Se Cond I$ Iimimun...
3 answers
For the following vector field calculate the fiuxpt surface $, F=<2y, 3x, z > , surface S:{x2+y2<9,<o,y<0,2 = 10} Canyou USC S[okL ; Divergence Theorem here? Justliy .
For the following vector field calculate the fiuxpt surface $, F=<2y, 3x, z > , surface S:{x2+y2<9,<o,y<0,2 = 10} Canyou USC S[okL ; Divergence Theorem here? Justliy ....
5 answers
Company that packages peanuts states that at maximum 6% of the peanut shells contain 300 peanuts were selected and 21 of them were empty.nuts_ At random_Withsignificance level of 1% , can the statement made by the company be accepted?State the null and alternative hypotheses:Calculate the critical valueDecide:With the same sample percentage of empty nuts and estimate the proportion of nuts with an error of less than 1%?0.95, what sample size would be needed
company that packages peanuts states that at maximum 6% of the peanut shells contain 300 peanuts were selected and 21 of them were empty. nuts_ At random_ With significance level of 1% , can the statement made by the company be accepted? State the null and alternative hypotheses: Calculate the criti...
5 answers
0 / 8 pts .(icoueu;;Question 3How many mL of J 2.50 M solution of Mg(OHJz do we need t0 neutralize 12.0 mol of HjPO4? (hint: You need to write and halarice the neutralization reaction)80.10"7,2050.10"30.10'20.10)45 020,0200110`
0 / 8 pts . (icoueu;; Question 3 How many mL of J 2.50 M solution of Mg(OHJz do we need t0 neutralize 12.0 mol of HjPO4? (hint: You need to write and halarice the neutralization reaction) 80.10" 7,20 50.10" 30.10' 20.10) 45 0 20,0 200110`...
5 answers
Find h =In terms of = and g'h(x)4()
Find h = In terms of = and g' h(x) 4()...
5 answers
12 a} L oithe function: Sxupod 1g6) = 2
12 a} L oithe function: Sxupod 1 g6) = 2...
5 answers
Answer -study conducted to test the claim that the mean IQ score of all HS students is more than 100. Use the information below to estimate the p-value for the studyn < sigma is unknown TR = 2.7910.01 20 0.005 < p 0.01 0.001 0.005 d) 0.002 0.01
Answer - study conducted to test the claim that the mean IQ score of all HS students is more than 100. Use the information below to estimate the p-value for the study n < sigma is unknown TR = 2.791 0.01 20 0.005 < p 0.01 0.001 0.005 d) 0.002 0.01...

-- 0.020421--