Course MT2002 Analysis

Solution 5

  1. (a) (2n/(n2+ 1)) =((2/n)/(1 + 1/n2)) rarrow 0/(1 + 0) = 0.
    (b) (n2 - 2n + 1)/(n2 + 2n + 1) = (1 - 2/n + 1/n2)/(1 + 2/n + 1/n2) and so (an) rarrow (1 - 0 + 0)/(1 + 0 + 0) = 1.
    (c) 3n/(2n + 3n) = 1/((2/3)n + 1) and so (an) rarrow 1/(0 + 1) = 1
    (d) sqrt(n + 1) - sqrtn = [(sqrt(n + 1) - sqrtn)(sqrt(n + 1) + sqrtn)]/(sqrt(n + 1) + sqrtn) = [(n + 1) - n]/(sqrt(n + 1) + sqrtn) = 1/(sqrt(n + 1) + sqrtn) and so the sequence rarrow 0 as n rarrow infinity.

  2. (a) Taking n = 1, 3, 7, 9, 13, 15, ... gives a subsequence (sqrt3/2 , 0 , sqrt3/2 , 0 , ... ) which does not converge. Hence the original sequence does not converge.
    (b) The subsequence of even terms rarrow 1 and the subsequence of odd terms rarrow -1 and so (since the limits of subsequences of a convergent sequence are the same as the limit of the sequence) the sequence is not convergent.

  3. 0 < n!/nn = (1.2.3. ... .n)/(n.n. ... .n) < 1/n. Since the constant sequence (0) rarrow 0 and the sequence (1/n) rarrow 0 the given sequence rarrow 0 by the "squeeze rule".

    Stirling's formula shows that n! grows approximately like sqrt(2p) nn+0.5e-n. For example, for n = 100, n! is about 0.93326 cross 10158 while Stirling's approximation is 0.93248 cross 10158. An even better approximation is obtained by replacing e-n by e-n+1/(12n). This gives an approximation of 0.93323 cross 10158 for 100!.

  4. an+1- an= 2an/(1+an) - 2an-1/(1+an-1) = 2(an- an-1)/[(1+an)(1+an-1)]
    Hence, provided an-1, an> -1, the sequence is monotonic.
    If a1 = 2 then a2 = 4/3 and so the sequence is monotonic decreasing and since it is clearly bounded below by 0, it is convergent.
    If a1 = 1/2 then a2 = 2/3 and so the sequence is monotonic increasing. It is easy to prove by induction that if a1 < 1 then an < 1 for all n and so the sequence is bounded above and hence convergent.
    In either case, the limit alpha must satisfy alpha = 2alpha/(1 + alpha) implies alpha = 1.

  5. Given epsilon > 0, choose N so that a2n is within epsilon of alpha for 2n > N and a2n-1 is within epsilon of alpha for 2n - 1 > N. Then all the terms of the sequence are eventually within epsilon of alpha and the sequence converges.

  6. The sequence (rn) is (1, 2, 1.5, 1.6, 1.625, 1.615, 1.619, 1.618, ... ) (approximately). Hence it is not monotonic -- though you might notice that the subsequences of odd and even terms are monotonic.
    rn+1 = fn+2/fn+1 = (fn+1 + fn)/fn+1 = 1 + fn/fn+1= 1 + 1/rn.
    So if the sequence (rn) converges to a limit alpha, alpha satisfies alpha = 1 + 1/alpha implies alpha2- alpha - 1 = 0 implies (if (alpha > 1) alpha = (sqrt5 + 1)/2.
    From the above, rn+2 = 1 + 1/rn+1 = 1 + 1/(1+1/rn) = 1 + rn/(rn+ 1) = (2rn+ 1)/(rn+ 1) as required.
    rn+2 - rn= (2rn + 1)/(rn + 1) - (2rn-2 + 1)/(rn-2 + 1) = 2(rn - rn-2)/[(rn + 1)(rn-2 + 1)].
    Hence by induction, since r3 - r1 > 0, the sequence of odd terms is monotonic increasing and since r4 - r2 < 0, the subsequence of even terms is monotonic decreasing. Hence, by question 4 above, the sequence converges to (sqrt5 + 1)/2.