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Incorrect answer

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If a/c > b/d, the sequence is increasing. If a/c < b/d, the sequence is decreasing. a,b,c,d are chosen so that $\\dfrac{a}{c} \\neq \\dfrac{b}{d}$.

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Correct answer

", "name": "u1"}, "tval": {"templateType": "anything", "group": "Ungrouped variables", "definition": "sqrt((1 / c) * ((10 ^ r * abs(b * c -(a * d))) / c -d))", "description": "", "name": "tval"}, "c": {"templateType": "anything", "group": "Ungrouped variables", "definition": "chcop(a,a)", "description": "", "name": "c"}}, "ungrouped_variables": ["a", "c", "b", "s1", "b1", "d", "r", "n", "tval", "mono", "t1", "t2", "t3", "u1", "u2", "v1", "v2", "w1"], "name": "When does a sequence get within $d$ of its limit?", "functions": {"chcop": {"type": "number", "language": "jme", "definition": "if(gcd(a,b)=1,b,chcop(a,random(1..20)))", "parameters": [["a", "number"], ["b", "number"]]}}, "parts": [{"customMarkingAlgorithm": "", "showCorrectAnswer": true, "prompt": "

Find the limit $\\ell$ of $\\{x_n\\}$. Input as a fraction or an integer.

\n

Limit $\\ell=$ [[0]]

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Find the least integer $N$ such that

\n

\\[\\left|{x_n -\\ell}\\right| < 10 ^ { -\\var{r}}, \\quad \\text{for } n \\geq N\\]

\n

Least $N=$ [[0]]

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{u1}

", "

{u2}

", "

{t3}

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Which one of the following describes $\\{x_n\\}$?

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Let

\n

\\[x_n=\\simplify[std]{({a}n^2+{b})/({c}n^2+{d})}, \\quad n=1,2,3, \\ldots\\]

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$x_n=\\frac{an^2+b}{cn^2+d}$. Find the least integer $N$ such that $\\left|x_n -\\frac{a}{c}\\right| < 10 ^{-r},\\;n\\geq N$, $2\\leq r \\leq 6$. Determine whether the sequence is increasing, decreasing or neither.

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a) 

\n

To begin with, the limit $\\ell$ is obtained by dividing top and bottom by $n^2$:

\n

\\[\\simplify[std]{({a}n^2+{b})/({c}n^2+{d})-{a}/{c}} = \\simplify[std]{({a}+{b}/n^2) /({c}+{d}/n^2)} \\to \\simplify[std]{({a})/({c})}\\] as $n \\to \\infty$, so $\\displaystyle \\ell= \\simplify[std]{{a}/{c}}$.

\n

b)

\n

To find the least $N$ such that all terms from the $N$th are less than $10^{\\var{-r}}$ from the limit we proceed as follows:

\n

\\begin{align}
\\left|\\simplify[std]{x_n -({a} / {c})}\\right| < 10 ^ { -\\var{r}} &\\iff \\left|\\simplify[std]{({a}n^2+{b})/({c}n^2+{d})-{a}/{c}}\\right| < 10 ^ { -\\var{r}} \\\\
&\\iff \\simplify[std]{abs({b*c-a*d})/({c^2}n^2+{c*d})} <10 ^ { -\\var{r}}
\\end{align}

\n

(We can get rid of the absolute value in the denominator as $\\simplify[std]{{c^2}n^2+{c*d}} \\gt 0$, $\\forall n=1,2,3,\\ldots$)

\n

Rearranging this last inequality by multiplying both sides by $(\\simplify[std]{{c^2}n^2+{c*d}})10^{\\var{r}}$ (this is positive and so the inequality does not reverse), we get:

\n

\\[\\simplify[std]{{c^2}n^2+{c*d}} > \\var{10^r*abs(b*c-a*d)} \\iff n^2 > \\frac{1}{\\var{c^2}}\\left(\\simplify[std]{{10^r*abs(b*c-a*d)}-{c*d}}\\right)=\\var{tval^2} \\iff n> \\var{tval}\\]

\n

Hence the least integer value is given by $N=\\var{N}$.

\n

c)

\n

Given $x_n = \\dfrac{an^2+b}{cn^2+d}, c \\gt 0, d\\gt 0$ it can be shown that $x_n \\leq x_{n+1} \\iff \\dfrac{b}{d} \\leq \\dfrac{a}{c}$. Here $\\dfrac{b}{d}=\\dfrac{\\var{b}}{\\var{d}}$ and $\\dfrac{a}{c}=\\dfrac{\\var{a}}{\\var{c}}$. Therefore the sequence will be increasing if $\\dfrac{\\var{b}}{\\var{d}} \\leq \\dfrac{\\var{a}}{\\var{c}} $ and decreasing if $\\dfrac{\\var{b}}{\\var{d}} \\geq \\dfrac{\\var{a}}{\\var{c}} $. Hence the sequence is $\\var{w1}$.

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