MT2002 Analysis

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(Exercises 6)

Exercises 5

  1. Use the arithmetic properties of convergent sequences to find the limits of the following sequences.
    (a) ( 2n/(n2+ 1) )
    (b) ( (n2- 2n + 1)/(n2+ 2n + 1) )
    (c) ( 3n/(2n+ 3n) )
    (d) (sqrt(n + 1) - sqrtn ) [Multiply top and bottom by sqrt(n + 1) + sqrtn]

    Solution to question 1

  2. By looking for suitable divergent subsequences, prove that the following sequences are divergent.
    (a) ( sin(np/3) )
    (b) ( (-1)nn/(2n+1) )

    Solution to question 2

  3. Prove that n!/nn< 1/n, and hence show that the sequence n!/nn converges to 0.
    [The Scottish mathematician (who was also the manager of a mine at Leadhills) James Stirling (1692 to 1770) showed that the sequence ( n!/(ennn+1/2) ) converges to sqrt(2p).]

    Solution to question 3

  4. A sequence (an) satisfies an+1= 2an/(1+an) for n = 1, 2, ...
    (a) If a1= 2, prove that the sequence is monotonic decreasing and bounded and find its limit.
    (b) If a1= 1/2 , prove that the sequence is monotonic increasing and bounded and find its limit.

    Solution to question 4

  5. Let (an) be a sequence such that the subsequence of even terms (a2n) and the subsequence of odd terms (a2n-1) both converge to the same limit alpha. Prove that (an) converges to alpha.

    Solution to question 5

  6. Let (fn) be the Fibonacci sequence (1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , ... ) satisfying f1 = f2 = 1 and fn+1= fn+ fn-1.
    Let (rn) be the sequence of ratios of successive Fibonacci numbers. So rn= fn+1/fn and (rn) = (1 , 2 , 3/2 , 5/3 , 8/5 , 13/8 , 21/13 , ...) . Is this sequence monotonic? [Use a calculator.]
    Prove that rn+1 = 1 + 1/rn and hence prove that if this sequence were convergent then its limit would be (sqrt5 + 1)/2. (This is the number that the Ancient Greeks called the Golden Ratio.)
    Prove that rn+2 = (2rn + 1)/(rn + 1).
    Hence show that the subsequence of odd terms and the subsequence of even terms are monotonic and bounded.
    Deduce that (rn) is convergent to the Golden Ratio.

    Solution to question 6

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JOC September 2002