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pc1 = p series convergent number 1

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gd2= geometric series divergent number 2 etc

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This question is about the comparison test for series.

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.

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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.

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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 diverge.

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You 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.

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You 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.

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What does the comparison test tell us about the series $\\sum t_k$?

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This series converges.

", "

This series diverges.

", "

It doesn't tell us anything.

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Suppose 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 convergent.

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Suppose 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.

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Suppose 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.

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Suppose 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.

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What does the comparison test tell us about the series $\\sum b_n$?

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This series converges.

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This series diverges.

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It doesn't tell us anything.

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Given the series

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$\\displaystyle\\sum_{k=\\var{start}}^\\infty$ {cexpression}

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This series converges.

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This series diverges.

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It doesn't tell us anything.

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The comparison test is as follows. Suppose that $0\\ge a_n\\ge b_n$ for sufficiently large $n$.

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• If $a_n$ diverges, then $b_n$ also diverges.

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• If $b_n$ converges, then $a_n$ also converges.

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Notice, 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'.

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a)

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The comparison test ensures us that our series converges (because a larger one does).

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The comparison test ensures us that our series diverges (because a smaller one does).

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b)

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The comparison test ensures us that our series converges (because a larger one does).

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The comparison test ensures us that our series diverges (because a smaller one does).

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c)

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In 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

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\$\\simplify{{a}/({d}k^{b}+{f}k+{g})<{a}/({d}k^{b})}\$

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and therefore our series is also convergent.

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In 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

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\$\\simplify{{a}/({d}k^{b}-{f}k-{g})>{a}/({d}k^{b})}\$

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and so can't use the comparison test to compare it to that series. But, for $k\\ge\\var{start}$ we definitely have

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\$\\simplify{{a}/({d}k^{b}-{f}k-{g})<{a}/({d}k^{b+1})}\$

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and $\\sum_{k=\\var{start}}^\\infty \\simplify{{a}/({d}k^{b+1})}$ is a convergent $p$-series. Therefore our series is also convergent.

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Notice that $0\\le \\cos^2(k)\\le 1$ and so for $k\\ge\\var{start}$ we definitely have

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\$\\simplify{{b}cos(k)^2/({d}k^{a})<={b}/({d}k^{a})}\$

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Also 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.

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Notice that $0\\le \\sin^2(k)\\le 1$ and so for $k\\ge\\var{start}$ we definitely have

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\$\\simplify{{b}sin(k)^2/({d}k^{a})<={b}/({d}k^{a})}\$

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Also 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.

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For $k\\ge\\var{start}$ we definitely have

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\$\\simplify{{a}ln(k)/({d}k)>{a}/({d}k)}\$

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and $\\sum_{k=\\var{start}}^\\infty \\simplify{{a}/({d}k)}$ 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.

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For $k\\ge\\var{start}$ we definitely have

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\$\\simplify{{a}/({d}ln(k))>{a}/({d}k)}\$

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and $\\sum_{k=\\var{start}}^\\infty \\simplify{{a}/({d}k)}$ 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.

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Given $\\simplify{{d}k^{b-1}/(sqrt(k^{2*b}+{a}))}$ we might realise that

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\$\\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}\$

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however, $\\sum \\simplify{{d}/k}$ 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:

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\$\\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)}\$

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Notice, $\\sum_{k=\\var{start}}^\\infty \\simplify{{d}/(sqrt({a+1})k)}$ 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.

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In 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

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\$\\simplify{{a}/({d}^k+{f})<{a}/({d}^k)}\$

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and therefore our series is also convergent.

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Notice that $0\\le \\cos^2(k)\\le 1$ and so for $k\\ge\\var{start}$ we definitely have

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\$\\simplify{{b}cos(k)^2/({d}{a}^k)<={b}/({d}{a}^k)}\$

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Also 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.

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Notice that $0\\le \\sin^2(k)\\le 1$ and so for $k\\ge\\var{start}$ we definitely have

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\$\\simplify{{b}sin(k)^2/({d}*{a}^k)<={b}/({d}*{a}^k)}\$

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Also 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.

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