EE 518 Homework 4 Due on Monday 2 00pm September 26 2016 Problem 1 n n Let an Please show that the sequence an is convergent and lim 0 where R is any real n n n number When 0 it is trivial that the sequence an is convergent and the limit is 0 When 0 n 1 n an 1 an n 1 n n 1 an 1 1 for any n 1 Therefore when n 1 the sequence an monotonically decrease an Since an 0 the sequence is bounded from below Therefore an converges Suppose limn a since an 1 an take limit on both sides we have n 1 a lim an 1 n lim an 0 a 0 n 1 n When 0 n n n n n n n 0 n n Use theorem of squeeze lim Problem 2 Find the limits of the following sequences q p p i 2 2 2 2 2 2 ii an n 1 n iii an n sin n n2 1 Page 1 of 5 EE 518 Homework 4 Due on Monday 2 00pm September 26 2016 i The sequence can be expressed as yy 2 2 2 4 8 2 2 2 Thus let 2 an 1 22n n 1 2 3 1 Since limn 2 2 n 1 then we have that lim an 2 n ii an n 1 n 1 n 1 n n 1 n n 1 n n 1 n Then lim an 0 n iii n sin n n2 1 0 we have that an Since sin n is bounded and limn n n2 1 lim an 0 n Problem 3 Let xn and yn be two number sequences suppose that for some fixed integer K we have yn xn 0 n K Please check whether following statements are true or false If true prove them If false provide at least one counter example i The convergence of yn implies the convergence of xn ii The divergence of xn implies the divergence of yn P P iii The convergence of n 0 xn implies the convergence of n 0 yn P P iv The divergence of n 0 xn implies the divergence of n 0 yn i False For example if yn 2 and xn 1 sin n yn converges but xn does not ii False Same example as in i iii False For example if xn 1 n2 and yn 1 n xn converges but yn does not iv True because of theorem 3 5 ii Problem 4 Discuss convergence or divergence for each of the following series Page 2 of 5 EE 518 Homework 4 i P ii P iii P iv P Due on Monday 2 00pm September 26 2016 n n 1 2n n 1 n 1 n 1 n sin n1 1 2n lnn 3 3n n 3n 1 an for a 0 i an 1 lim n an n lim n 1 2n 1 n 2n lim n n 1 1 1 2n 2 By ratio test rule the series is convergent ii lim n sin n sin n1 1 lim 1 6 0 n n n1 By theorem 3 2 the series is divergent P 1 P 1 iii Since by theorem 3 4 then n 1 n 1 2n is also divergent In addition n n limn ln3n 3 0 Therefore the whole series is divergent lnn 3 3n 0 n and iv By root test 1 lim sup an n lim sup n n 1 n a a 1 3n 1 3 a 0 Therefore the series is convergent Problem 5 1 ln x ex h sin x and g x x3 1 Let f x x 5ex 1 x2 cos x 1 i Find the following limits x2 4 i lim f x lim g x x 0 x 0 ii lim g f x x 0 i 1 lim f x 1 lim g x 1 5 2 x 0 1 4 x 0 ii lim g f x 5 5 x 0 Problem 6 Find the continuous intervals and discontinuities of following functions Indicate what kind of discontinuities they are i f x x2 x x x2 1 Page 3 of 5 EE 518 Homework 4 sin x 1 ex x 4 1 ii f x 2 x 4 2 x Due on Monday 2 00pm September 26 2016 x 0 x 0 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 2 Problem 7 Let Sn n X 1 and write a MATLAB function that finds the minimum required n for Sn to reach the sum k2 k 1 2 limit within a given positive error that is less than 1 The input argument of the function should be 6 error and the output should be n such that Sn 2 error 6 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 5 EE 518 Homework 4 Due on Monday 2 00pm September 26 2016 error 0 1 0 05 0 01 0 001 0 0001 n 10 20 100 1000 10000 Page 5 of 5