EE 518 : Homework #4 Due on Monday 2:00pm, September 26, 2016Problem 1Let an=αnn!. Please show that the sequence {an} is convergent, and limn→∞αnn!= 0, where α ∈ R is any realnumber.When α = 0, it is trivial that the sequence {an} is convergent and the limit is 0. When α > 0,an+1an=αn+1(n + 1)!,αnn!=αn + 1an+1an< 1, for any n > α − 1. Therefore, when n > α − 1, the sequence {an} monotonically decrease.Since an> 0, the sequence is bounded from below. Therefore {an} converges. Suppose limn→∞= a, sincean+1=αn + 1an, take limit on both sides, we havea = limn→∞an+1=αn + 1limn→∞an= 0 · a = 0.When α < 0,−|α|nn!≤αnn!≤|α|nn!Use theorem of squeeze, limn→∞αnn!= 0.Problem 2Find the limits of the following sequences.(i)√2,p2√2,q2p2√2, . . .(ii) an= (√n + 1 −√n)(iii) an=n sin n!n2+1Page 1 of 5EE 518 : Homework #4 Due on Monday 2:00pm, September 26, 2016(i) The sequence can be expressed as yy2√2,24√2,28√2, . . .Thus, letan=2212n, n = 1, 2, 3, ...Since limn→∞212n= 1, then we have that:limn→∞an= 2(ii)an=√n + 1 −√n = (√n + 1 −√n) ·√n + 1 +√n√n + 1 +√n=1√n + 1 +√nThen,limn→∞an= 0(iii)an=n sin n!n2+ 1Since, sin n! is bounded and limn→∞nn2+1= 0, we have that:limn→∞an= 0Problem 3Let {xn} and {yn} be two number sequences, suppose that for some fixed integer K, we haveyn≥ xn≥ 0, ∀n ≥ K.Please check whether following statements are true or false. If true, prove them. If false, provide at leastone counter-example.(i) The convergence of {yn} implies the convergence of {xn};(ii) The divergence of {xn} implies the divergence of {yn};(iii) The convergence ofP∞n=0xnimplies the convergence ofP∞n=0yn;(iv) The divergence ofP∞n=0xnimplies the divergence ofP∞n=0yn.(i) False. For example, if yn= 2 and xn= 1 + sin(n), ynconverges, but xndoes not.(ii) False. Same example as in (i)(iii) False. For example, if xn=1n2and yn=1n, xnconverges, but yndoes not.(iv) True, because of theorem 3.5(ii).Problem 4Discuss convergence or divergence for each of the following series:Page 2 of 5EE 518 : Homework #4 Due on Monday 2:00pm, September 26, 2016(i)P∞n=1n2n;(ii)P∞n=1n sin1n;(iii)P∞n=112n+lnn33n(iv)P∞n=1n3n+1an, for a > 0(i)limn→∞an+1an= limn→∞n+12n+1n2n= limn→∞n + 12n=12< 1By ratio test rule, the series is convergent.(ii)limn→∞n sin1n= limn→∞sin1n1n= 1 6= 0By theorem 3.2, the series is divergent.(iii) SinceP∞n=11nby theorem 3.4, thenP∞n=112nis also divergent. In addition,lnn33n> 0 ∀n andlimn→∞lnn33n= 0. Therefore, the whole series is divergent.(iv) By root test,limn→∞sup |an|1n= limn→∞sup(n3n + 1)a=13a< 1, ∀a > 0Therefore, the series is convergent.Problem 5Let f(x) =1 + ln(x + ex)sin xx1x2+4and g(x) =hx3+ 1 5exx2cos x−1iFind the following limits:(i) limx→0f(x), limx→0g(x)(ii) limx→0(g · f)(x)(i)limx→0f(x) =1114, limx→0g(x) = [1, 5, −2](ii)limx→0(g · f)(x) = 5.5Problem 6Find the continuous intervals and discontinuities of following functions. Indicate what kind of discontinuitiesthey are.(i) f(x) =x2− x|x|(x2− 1)Page 3 of 5EE 518 : Homework #4 Due on Monday 2:00pm, September 26, 2016(ii) f(x) =sin xx− ex+14, x < 012, x = 0√x + 4 − 2x, x > 0(i) The continuous intervals are (−∞, −1), (−1, 0), (0, 1), (1, ∞)x = −1 is a second kind discontinuity f (−1−) and f(−1+) do not exist.x = 0 is a first kind discontinuity f (0−) = −1 and f(0+) = 1.x = 1 is a first kind discontinuity f (1−) = 1/2 and f(1+) = 1/2.(ii) Continuous intervals: (−∞, 0), (0, ∞)x = 0 is a first kind discontinuity f (0−) = f(0+)1/4 but f(0) = 1/2Problem 7Let Sn=nXk=11k2and write a MATLAB function that finds the minimum required n for Snto reach the sumlimitπ26within a given positive error that is less than 1. The input argument of the function should beerror and the output should be n, such that:Sn−π26≤ error.With your function find out the values of n when error is 0.1, 0.05, 0.01, 0.001 and 0.0001 respectively.(Hint: Write a function that computes Sn, then write the function calling this function to find n.)Page 4 of 5EE 518 : Homework #4 Due on Monday 2:00pm, September 26, 2016error n0.1 100.05 200.01 1000.001 10000.0001 10000Page 5 of