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    40 Problems : Sequences, Series, Convergence, Divergence and Limits

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    1. For each of the sequences whose nth term is given by the formula below (so of course n takes successively the positive integer values 1,2,3...), does it have a limit as n tends to infinity?
    In each case, briefly explain your answer including justification for the value of the limit (if it exists)

    a) (1/3)ⁿ
    b) 2ⁿ
    c) (1.01) ⁿ
    d) (0.99) ⁿ
    e) 1/10ⁿ
    f) 2‾ ⁿ
    g) (-(1/5)) ⁿ
    h) (-5) ⁿ
    i) 5ⁿ/6ⁿ
    j) 2³ⁿ/3²ⁿ

    and 1-xⁿ/1-x when,

    k) x = 2/3
    l) x = -(1/2)
    m) x = 2
    n) x = -3
    o) x = 1
    p) x = -1

    2. Determine the limits of the sequences whose nth terms are given by the following formulae. Briefly explain your answers.

    a) 2n+1/3n-1
    b) n-5n+1/n+n-1
    c) 3n-1/n+n+1
    d) 5n/10n-1
    e) (n+n)-n/(5n-1)
    f) (2n/2n-1)-(n/n+1)

    formulate a general rule for the limit of any sequence whose nth term has the form P(n)/Q(n)
    where P and Q are polynomial functions. The rule will need to take account of the degree of each of these polynomials(beginning with the highest power of n that occurs - the notation deg(P) is often used for the degree of (P). it will also need to refer to the leading coefficient of each of these polynomials.

    Determine the limit(where there is one ) for the sequences whose nth terms are given by the following formulae. Give brief justification.
    g) n/3
    h) 2ⁿ/10n-1
    i) n/10ⁿ
    j) n^10/(1.1) ⁿ
    k) n^-5/(-5) ‾ⁿ

    formulate general rules for the sequences of this type.

    3. Determine whether these series are convergent or divergent

    a) ∑ (3r + 5)
    r = 1

    b) ∑ (20-3s)
    s = 1

    c) ∑ (1/2)^r
    r = 1

    d) ∑ (3/2)^(r+1)
    r = 1

    e) ∑ (-(1/3))^k-1
    k = 1

    f) ∑ 1/r(r+1)
    r = 1

    g) ∑ 1/r(r+2)
    r = 1

    h) ∑ log(r+1/r)
    r = 1

    i) ∑ r/(r+1)!
    r = 1

    j) ∑ r-1/2^(r+1)
    r = 1

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    Solution Summary

    Forty Problems involving Sequences, Series, Convergence, Divergence and Limits are solved. The solution is detailed and well presented.