Porisma

Discussion questions · Section 2 of 3

Questions 7–12 in this collection

Bound the tail of a convergent geometric series · Compare arithmetic and geometric partial sums · +4 more

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  1. 7

    The series 1+e−x+e−2x+⋯1+e^{-x}+e^{-2x}+\cdots has a sum to infinity. Show that x>0x>0. When x=10x=10, write SnS_n and SS, the finite and infinite sums. Show that S−Sn<S/10100S-S_n<S/10^{100} requires n>10ln⁡10n>10\ln10.

    CJC · Discussion · Question 7

  2. 8

    The seventh, third and first AP terms (nonzero common difference) are respectively the first three GP terms. Show that the geometric series converges. The seventh AP term is 33. Find the smallest nn for which the first nn AP terms exceed the first nn GP terms by at least 100100.

    [2008/AJC/II/2(modified)]

    CJC · Discussion · Question 8

  3. 9

    Find (i) ∑k=0n(k+n+1)\sum_{k=0}^n(k+n+1) in terms of nn; (ii) ∑r=1N(4r+2−r)\sum_{r=1}^N(4r+2^{-r}) in terms of NN; (iii) ∑k=130(a2k+ka)\sum_{k=1}^{30}(a^{2k}+ka) in terms of aa.

    CJC · Discussion · Question 9

  4. 10

    Given ∑r=1nr(r+2)=n(n+1)(2n+7)/6\sum_{r=1}^nr(r+2)=n(n+1)(2n+7)/6, find ∑r=n2nr(r+2)\sum_{r=n}^{2n}r(r+2).

    CJC · Discussion · Question 10

  5. 11

    Given ∑r=1n7r+10r(r+1)(r+2)=6−(5n+1+2n+2)\sum_{r=1}^n\dfrac{7r+10}{r(r+1)(r+2)}=6-\left(\dfrac5{n+1}+\dfrac2{n+2}\right), deduce the exact value of ∑r=22∞21r+30r(r+1)(r+2)\sum_{r=22}^\infty\dfrac{21r+30}{r(r+1)(r+2)}.

    [2007/HCI/I/7(modified)]

    CJC · Discussion · Question 11

  6. 12

    Given ∑r=2n3r2+1(r2−1)3=12[98−1n3−1(n+1)3]\sum_{r=2}^n\dfrac{3r^2+1}{(r^2-1)^3}=\dfrac12\left[\dfrac98-\dfrac1{n^3}-\dfrac1{(n+1)^3}\right], find ∑r=3n3(r−1)2+1[(r−1)2−1]3\sum_{r=3}^n\dfrac{3(r-1)^2+1}{[(r-1)^2-1]^3} in terms of nn. Hence find its infinite sum.

    [2009/SAJC/II/3(modified)]

    CJC · Discussion · Question 12

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