I need a help to solve these four calculus questions.

1.

(2 marks) Determine whether the series below converge or diverge and select the appropriate test for each. 0) i0: k k=2 5162 2 diverges What test should be used to give this result? Comparison test v (ll) 0 k+75 2' k!) k=1 I converges What test should be used to give this result? Ratio test I (III) f: (1)"6 k=1 5 3": diverges I What test should be used to give this result? k-th term test I (2 marks) Consider the series E 1 \"=1 +n7 In order to address the convergence/divergence of this series we will use the comparison test. From the options below, select the appropriate statement to begin the argument. 4 +n7 C, We will find g('n.) such that O S g('n.) S xx, We will find g(n) such that 0 S S g('n.) , for all n 2 1, to prove that the above series diverges. for all 'n. 2 1, to prove that the above series diverges. 4 +n7 ' 1 g g(n) , for all n 2 1, to prove that the above series oonverges. +n7 A, We will nd g(n) such that 0 S 9(n) g 0 We will nd g(n) such that 0 S 4 +n7 , for all 'n. 2 1, to prove that the above series converges. A suitable expression for g (n) is 9h = _..:' h =19 (2 marks) Determine the limiting behaviour of the following sequences as n - co. If possible, simplify your answer. For OO write infinity - OO write -infinity TT write Pi does not exist (other than co or -co ) write undefined i) An (1+ 3 n You may use the standard result lim 1+ 1 n n-too n = e = exp(1) ii) In (9n2+ 7) -+ V2n + 4 iii) -1 -3n-+3n+1 COS 6n2 + 4( 2 marks ) Consider the series 7k (k + 1)! i) Find the values of A and B in the following identity. A B 7k K ! (k + 1)! (k + 1)! A = and B = ii) Now, using the technique of the telescoping series, we can evaluate the n' partial sum, sn . 7k Sn = C = 7+ (k + 1)! DL where C = is a constant and D= is a function of n. 7 k iii) Hence, we can conclude that (k + 1)! converges and we can evaluate the sum to infinity. IM8 IM8 7k lim Sn = (k + 1)! n-too