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pc1 = p series convergent number 1
\ngd2= geometric series divergent number 2 etc
\n", "definition": "if(ccondiv='div',random('pd1','pd2','pd3'),random('pc1','pc2','pc3','pc4','gc1','gc2','gc3'))", "name": "cseed", "templateType": "anything"}, "pc2": {"group": "Ungrouped variables", "description": "", "definition": "'\\$\\\\displaystyle\\\\simplify{{a}/({d}k^{b}-{f}k-{g})}\\$'", "name": "pc2", "templateType": "anything"}, "pd2": {"group": "Ungrouped variables", "description": "", "definition": "'\\$\\\\displaystyle\\\\simplify{{a}/({d}*ln(k))}\\$'", "name": "pd2", "templateType": "anything"}, "a": {"group": "Ungrouped variables", "description": "", "definition": "coeff[0]", "name": "a", "templateType": "anything"}, "gc3": {"group": "Ungrouped variables", "description": "", "definition": "'\\$\\\\displaystyle\\\\simplify{{b}*sin(k)^2/({d}*{a}^k)}\\$'", "name": "gc3", "templateType": "anything"}, "limit": {"group": "Ungrouped variables", "description": "", "definition": "if(seed=1,random(1/9,1/8,1/7,1/5,1/4,1/3,1/2,1/10,1/100),if(seed=0,random(10/9,11/8,8/7,7/5,5/4,4/3,5/2,2,17/10,102/100),1))", "name": "limit", "templateType": "anything"}, "coeff": {"group": "Ungrouped variables", "description": "", "definition": "shuffle(2..12)[0..6]", "name": "coeff", "templateType": "anything"}, "start": {"group": "Ungrouped variables", "description": "", "definition": "coeff[5]", "name": "start", "templateType": "anything"}, "pc1": {"group": "Ungrouped variables", "description": "", "definition": "'\\$\\\\displaystyle\\\\simplify{{a}/({d}k^{b}+{f}k+{g})}\\$'", "name": "pc1", "templateType": "anything"}, "b": {"group": "Ungrouped variables", "description": "", "definition": "coeff[1]", "name": "b", "templateType": "anything"}, "gc1": {"group": "Ungrouped variables", "description": "", "definition": "'\\$\\\\displaystyle\\\\simplify[fractionNumbers]{{a}/({d}^k+{f})}\\$'", "name": "gc1", "templateType": "anything"}, "pd3": {"group": "Ungrouped variables", "description": "", "definition": "'\\$\\\\displaystyle\\\\simplify{{d}k^{b-1}/(sqrt(k^{2*b}+{a}))}\\$'", "name": "pd3", "templateType": "anything"}, "f": {"group": "Ungrouped variables", "description": "", "definition": "coeff[3]", "name": "f", "templateType": "anything"}, "pd1": {"group": "Ungrouped variables", "description": "", "definition": "'\\$\\\\displaystyle\\\\simplify{{a}ln(k)/({d}*k)}\\$'", "name": "pd1", "templateType": "anything"}, "pc3": {"group": "Ungrouped variables", "description": "", "definition": "'\\$\\\\displaystyle\\\\simplify{{b}*cos(k)^2/({d}k^{a})}\\$'", "name": "pc3", "templateType": "anything"}, "seed": {"group": "Ungrouped variables", "description": "", "definition": "random(0..3)", "name": "seed", "templateType": "anything"}, "d": {"group": "Ungrouped variables", "description": "", "definition": "coeff[2]", "name": "d", "templateType": "anything"}, "ccondiv": {"group": "Ungrouped variables", "description": "", "definition": "random('con','div')", "name": "ccondiv", "templateType": "anything"}, "g": {"group": "Ungrouped variables", "description": "", "definition": "coeff[4]", "name": "g", "templateType": "anything"}, "gc2": {"group": "Ungrouped variables", "description": "", "definition": "'\\$\\\\displaystyle\\\\simplify{{b}*cos(k)^2/({d}*{a}^k)}\\$'", "name": "gc2", "templateType": "anything"}}, "rulesets": {}, "variable_groups": [], "functions": {}, "ungrouped_variables": ["seed", "percent", "limit", "ccondiv", "cseed", "cexpression", "coeff", "a", "b", "d", "f", "g", "start", "pc1", "pc2", "pc3", "pc4", "pd1", "pd2", "pd3", "gc1", "gc2", "gc3"], "statement": "This question is about the comparison test for series.
", "metadata": {"description": "Test whether a student knows the comparison test of a series, and how to use it. Series include those that the comparison test is inconclusive for.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "parts": [{"displayType": "dropdownlist", "minMarks": 0, "shuffleChoices": false, "prompt": "You have a series $\\sum t_k$ with positive terms. Each term in the series is greater than or equal to the corresponding term in another series of positive terms which is actually known to converge.
\nYou have a series $\\sum t_k$ with positive terms. Each term in the series is greater than or equal to the corresponding term in another series of positive terms which is actually known to diverge.
\nYou have a series $\\sum t_k$ with positive terms. Each term in the series is less than or equal to the corresponding term in another series of positive terms which is actually known to converge.
\nYou have a series $\\sum t_k$ with positive terms. Each term in the series is less than or equal to the corresponding term in another series of positive terms which is actually known to diverge.
\nWhat does the comparison test tell us about the series $\\sum t_k$?
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\nSuppose you have two series, $\\sum a_n$ and $\\sum b_n$, where $a,b>0$, $a_n\\ge b_n$ for all $n$, and $\\sum a_n$ is divergent.
\nSuppose you have two series, $\\sum a_n$ and $\\sum b_n$, where $a,b>0$, $a_n\\le b_n$ for all $n$, and $\\sum a_n$ is convergent.
\nSuppose you have two series, $\\sum a_n$ and $\\sum b_n$, where $a,b>0$, $a_n\\le b_n$ for all $n$, and $\\sum a_n$ is divergent.
\nWhat does the comparison test tell us about the series $\\sum b_n$?
", "displayColumns": 0, "variableReplacements": [], "type": "1_n_2", "showFeedbackIcon": true, "scripts": {}, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "marks": 0, "maxMarks": 0, "distractors": ["", "", ""], "choices": ["This series converges.
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\n$\\displaystyle\\sum_{k=\\var{start}}^\\infty$ {cexpression}
\nWhat does the comparison test tell us about this series?
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"], "matrix": ["if(ccondiv='con',1,0)", "if(ccondiv='div',1,0)", "0"]}], "preamble": {"js": "", "css": ""}, "advice": "The comparison test is as follows. Suppose that $0\\ge a_n\\ge b_n$ for sufficiently large $n$.
\n• If $a_n$ diverges, then $b_n$ also diverges.
\n• If $b_n$ converges, then $a_n$ also converges.
\nNotice, if $a_n$ converges or if $b_n$ diverges the test doesn't say anything, in these cases some might say the test is 'inconclusive' or 'fails'.
\n\na)
\nThe comparison test ensures us that our series converges (because a larger one does).
\nThe comparison test ensures us that our series diverges (because a smaller one does).
\nThe comparison test doesn't tell us anything about this situation.
\n\nb)
\nThe comparison test ensures us that our series converges (because a larger one does).
\nThe comparison test ensures us that our series diverges (because a smaller one does).
\nThe comparison test doesn't tell us anything about this situation.
\n\nc)
\nIn the denominator of {cexpression} the dominant term is $\\var{d}k^\\var{b}$, so we will compare our series with $\\sum_{k=\\var{start}}^\\infty\\frac{\\var{a}}{\\var{d}k^\\var{b}}$ which is a $p$-series with $p=\\var{b}$ and hence is convergent. Now for $k\\ge\\var{start}$ we definitely have
\n\\[\\simplify{{a}/({d}k^{b}+{f}k+{g})<{a}/({d}k^{b})}\\]
\nand therefore our series is also convergent.
\nIn the denominator of {cexpression} the dominant term is $\\var{d}k^\\var{b}$, so we will compare our series with $\\sum_{k=\\var{start}}^\\infty\\frac{\\var{a}}{\\var{d}k^\\var{b}}$ which is a $p$-series with $p=\\var{b}$ and hence is convergent. However, for $k\\ge\\var{start}$ we have
\n\\[\\simplify{{a}/({d}k^{b}-{f}k-{g})>{a}/({d}k^{b})}\\]
\nand so can't use the comparison test to compare it to that series. But, for $k\\ge\\var{start}$ we definitely have
\n\\[\\simplify{{a}/({d}k^{b}-{f}k-{g})<{a}/({d}k^{b+1})}\\]
\nand $\\sum_{k=\\var{start}}^\\infty \\simplify{{a}/({d}k^{b+1})}$ is a convergent $p$-series. Therefore our series is also convergent.
\nNotice that $0\\le \\cos^2(k)\\le 1$ and so for $k\\ge\\var{start}$ we definitely have
\n\\[\\simplify{{b}cos(k)^2/({d}k^{a})<={b}/({d}k^{a})}\\]
\nAlso notice, $\\sum_{k=\\var{start}}^\\infty \\frac{\\var{b}}{\\var{d}k^{\\var{a}}}$ is a convergent $p$-series with $p=\\var{a}$. Therefore our series is also convergent.
\nNotice that $0\\le \\sin^2(k)\\le 1$ and so for $k\\ge\\var{start}$ we definitely have
\n\\[\\simplify{{b}sin(k)^2/({d}k^{a})<={b}/({d}k^{a})}\\]
\nAlso notice, $\\sum_{k=\\var{start}}^\\infty \\frac{\\var{b}}{\\var{d}k^{\\var{a}}}$ is a convergent $p$-series with $p=\\var{a}$. Therefore our series is also convergent.
\nFor $k\\ge\\var{start}$ we definitely have
\n\\[\\simplify{{a}ln(k)/({d}k)>{a}/({d}k)}\\]
\nand $\\sum_{k=\\var{start}}^\\infty \\simplify{{a}/({d}k)}$ is a divergent series (it is a scalar multiple of the harmonic series or a $p$-series with $p=1$). Therefore our series is also divergent.
\nFor $k\\ge\\var{start}$ we definitely have
\n\\[\\simplify{{a}/({d}ln(k))>{a}/({d}k)}\\]
\nand $\\sum_{k=\\var{start}}^\\infty \\simplify{{a}/({d}k)}$ is a divergent series (it is a scalar multiple of the harmonic series or a $p$-series with $p=1$). Therefore our series is also divergent.
\nGiven $\\simplify{{d}k^{b-1}/(sqrt(k^{2*b}+{a}))}$ we might realise that
\n\\[\\simplify{{d}k^{b-1}/(sqrt(k^{2*b}+{a}))<{d}k^{b-1}/(sqrt(k^{2*b}))={d}k^{b-1}/(k^{b})={d}/k}\\]
\nhowever, $\\sum \\simplify{{d}/k}$ is divergent (it is a scalar multiple of the harmonic series or a $p$-series with $p=1$) and given the direction of the inequality we can't use the comparison test to test these two series. But, lets try something else:
\n\\[\\simplify{{d}k^{b-1}/(sqrt(k^{2*b}+{a}))>{d}k^{b-1}/(sqrt(k^{2*b}+{a}k^{2*b}))={d}k^{b-1}/(sqrt({a+1}k^{2*b}))={d}k^{b-1}/(sqrt({a+1})k^{b})={d}/(sqrt({a+1})k)}\\]
\nNotice, $\\sum_{k=\\var{start}}^\\infty \\simplify{{d}/(sqrt({a+1})k)}$ is a divergent series (it is a scalar multiple of the harmonic series or a $p$-series with $p=1$) and so our series is too.
\nIn the denominator of {cexpression} the dominant term is $\\var{d}^k$, so we will compare our series with $\\sum_{k=\\var{start}}^\\infty\\frac{\\var{a}}{\\var{d}^k}$ which is the same as $\\sum_{k=\\var{start}}^\\infty\\var{a}\\left(\\frac{1}{\\var{d}}\\right)^k$ and so is a convergent geometric series with common ratio $r=\\frac{1}{\\var{d}}$. Now for $k\\ge\\var{start}$ we definitely have
\n\\[\\simplify{{a}/({d}^k+{f})<{a}/({d}^k)}\\]
\nand therefore our series is also convergent.
\nNotice that $0\\le \\cos^2(k)\\le 1$ and so for $k\\ge\\var{start}$ we definitely have
\n\\[\\simplify{{b}cos(k)^2/({d}{a}^k)<={b}/({d}{a}^k)}\\]
\nAlso notice, $\\sum_{k=\\var{start}}^\\infty \\frac{\\var{b}}{\\var{d}\\times\\var{a}^k}$ is a convergent $p$-series with with $p=\\var{a}$. Therefore our series is also convergent.
\nNotice that $0\\le \\sin^2(k)\\le 1$ and so for $k\\ge\\var{start}$ we definitely have
\n\\[\\simplify{{b}sin(k)^2/({d}*{a}^k)<={b}/({d}*{a}^k)}\\]
\nAlso notice, $\\sum_{k=\\var{start}}^\\infty \\frac{\\var{b}}{\\var{d}\\times\\var{a}^k}$ is a convergent $p$-series with with $p=\\var{a}$. Therefore our series is also convergent.
\n", "name": "Series: comparison test", "variablesTest": {"condition": "", "maxRuns": 100}, "type": "question", "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}]}]}], "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}]}