// Numbas version: exam_results_page_options {"name": "Series: p series", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"statement": "

You are given the series

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\$\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{{num}/({den}*k^{p})}.\$

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The $p$-series $\\sum_{n=1}^\\infty \\frac{1}{n^p}$ is convergent if $p>1$ and divergent if $p\\le 1$.

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Our series is $\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{{num}/({den}*k^{p})}$, which looks similar with a $p$-value of $\\var[fractionNumbers]{p}$. In fact:

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\\\begin{align}\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{{num}/({den}*k^{p})}&=\\simplify[fractionNumbers]{{num}/{den}}\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{1/(k^{p})}\\\\&=\\simplify[fractionNumbers]{{num}/{den}}\\left(-1+\\sum_{k=1}^\\infty \\simplify[fractionNumbers]{1/k^{p}}\\right)\\end{align}\

\n

\\\begin{align}\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{{num}/({den}*k^{p})}&=\\simplify[fractionNumbers]{{num}/{den}}\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{1/(k^{p})}\\\\&=\\simplify[fractionNumbers]{{num}/{den}}\\left(-1-\\frac{1}{2^\\var[fractionNumbers]{p}}+\\sum_{k=1}^\\infty \\simplify[fractionNumbers]{1/k^{p}}\\right)\\end{align}\

\n

\\\begin{align}\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{{num}/({den}*k^{p})}&=\\simplify[fractionNumbers]{{num}/{den}}\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{1/(k^{p})}\\\\&=\\simplify[fractionNumbers]{{num}/{den}}\\left(-1-\\frac{1}{2^\\var[fractionNumbers]{p}}-\\frac{1}{3^\\var[fractionNumbers]{p}}+\\sum_{k=1}^\\infty \\simplify[fractionNumbers]{1/k^{p}}\\right)\\end{align}\

\n

\\\begin{align}\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{{num}/({den}*k^{p})}&=\\simplify[fractionNumbers]{{num}/{den}}\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{1/(k^{p})}\\\\&=\\simplify[fractionNumbers]{{num}/{den}}\\left(-1-\\frac{1}{2^\\var[fractionNumbers]{p}}-\\frac{1}{3^\\var[fractionNumbers]{p}}-\\frac{1}{4^\\var[fractionNumbers]{p}}+\\sum_{k=1}^\\infty \\simplify[fractionNumbers]{1/k^{p}}\\right)\\end{align}\

\n

\\\begin{align}\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{{num}/({den}*k^{p})}&=\\simplify[fractionNumbers]{{num}/{den}}\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{1/(k^{p})}\\\\&=\\simplify[fractionNumbers]{{num}/{den}}\\left(-1-\\frac{1}{2^\\var[fractionNumbers]{p}}-\\frac{1}{3^\\var[fractionNumbers]{p}}-\\frac{1}{4^\\var[fractionNumbers]{p}}-\\frac{1}{5^\\var[fractionNumbers]{p}}+\\sum_{k=1}^\\infty \\simplify[fractionNumbers]{1/k^{p}}\\right)\\end{align}\

\n

\\\begin{align}\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{{num}/({den}*k^{p})}&=\\simplify[fractionNumbers]{{num}/{den}}\\sum_{k=\\var{start}}^\\infty \\simplify[fractionNumbers]{1/(k^{p})}\\\\&=\\simplify[fractionNumbers]{{num}/{den}}\\left(-1-\\frac{1}{2^\\var[fractionNumbers]{p}}-\\frac{1}{3^\\var[fractionNumbers]{p}}-\\frac{1}{4^\\var[fractionNumbers]{p}}-\\frac{1}{5^\\var[fractionNumbers]{p}}-\\frac{1}{6^\\var[fractionNumbers]{p}}+\\sum_{k=1}^\\infty \\simplify[fractionNumbers]{1/k^{p}}\\right)\\end{align}\

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and so the convergence/divergence of our series depends on the convergence/divergence of the related $p$-series. Note that adding or subtracting a finite number of terms to a series will not change whether it converges or diverges (such as our missing first term terms), nor will multiplying or dividing the series by a non-zero constant (such as $\\frac{\\var{num}}{\\var{den}}$).

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So we have a convergent divergent series that you could say is really a $p$-series with $p=\\var[fractionNumbers]{p}$.

\n

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This question tests to see if students can recognise a $p$-series and based on its $p$-value determine if it is convergent or divergent.

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This is a [[0]] [[1]].

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convergent

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divergent

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p series

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geometric series

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arithmetic series

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alternating series

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Maclaurin series

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Taylor series

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