You are given the series
\n\\[\\sum_{k=\\var{start}}^\\infty\\var{a}\\left(\\var[fractionNumbers]{r}\\right)^k.\\]
", "metadata": {"description": "This question tests to see if students can recognise a geometric series and based on its common ratio determine if it is convergent or divergent.
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", "divergent
"], "matrix": ["if(abs(r)<1,1,0)", "if(abs(r)>=1,1,0)"]}, {"displayType": "dropdownlist", "minMarks": 0, "shuffleChoices": true, "displayColumns": 0, "variableReplacements": [], "type": "1_n_2", "showFeedbackIcon": true, "scripts": {}, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "marks": 0, "maxMarks": 0, "distractors": ["", "", "", "", "", ""], "choices": ["p series
", "geometric series
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"], "matrix": ["0", "1", 0, "if(r<0,1,0)", 0, 0]}], "showFeedbackIcon": true, "scripts": {}}], "preamble": {"js": "", "css": ""}, "advice": "A geometric series $\\sum_{n=0}^\\infty ar^n$ is convegent if $|r|<1$ with a sum of $\\frac{a}{1-r}$, and is divergent if $|r|\\ge1$.
\nOur series is $\\sum_{k=\\var{start}}^\\infty\\var{a}\\left(\\var[fractionNumbers]{r}\\right)^k$, which looks similar with an $r$ value of $\\var[fractionNumbers]{r}$. In fact, if we are worried that our series starts at $k=3$ instead of at $n=0$ notice:
\n\n$\\displaystyle\\sum_{k=\\var{start}}^\\infty \\var{a}\\left(\\var[fractionNumbers]{r}\\right)^k=\\sum_{k=\\var{start}}^\\infty \\var{a}\\left(\\var[fractionNumbers]{r}\\right)^{\\var{start}}\\left(\\var[fractionNumbers]{r}\\right)^{k-\\var{start}}$
\nwe now make the substitution $n=k-\\var{start}$ (which means when $k=\\var{start}$, $n=0$) and we have
\n$\\displaystyle\\sum_{n=0}^\\infty \\var{a}\\left(\\var[fractionNumbers]{r}\\right)^{\\var{start}}\\left(\\var[fractionNumbers]{r}\\right)^{n}$
\n\n\nand so even though our series seemed to start later on, it is still a geometric series with $a=\\var{a}\\left(\\var[fractionNumbers]{r}\\right)^{\\var{start}}$ and $r=\\var[fractionNumbers]{r}$. Also, note that adding or subtracting a finite number of terms to a series will not change whether it converges or diverges, nor will multiplying or dividing the series by a non-zero constant.
\nSo we have a convergent divergent geometric series with common ratio $r=\\var[fractionNumbers]{r}$.
\n\nSince this common ratio is negative, each time we multiply by it we alternate the sign of the term, that is, this series is also an alternating series. If we wanted to, we could pull the common factor of $-1$ out and write our series as a positive number multiplied by $(-1)^n$:
\n$\\displaystyle\\sum_{n=0}^\\infty (-1)^n\\var{abs(a)}\\left(\\var[fractionNumbers]{abs(r)}\\right)^{\\var{start}}\\left(\\var[fractionNumbers]{abs(r)}\\right)^{n}$
\nThe alternating series $\\sum_{n=0}^\\infty (-1)^n b_n$ where $b_n>0$, converges if $b_{n+1}\\le b_n$ for all $n$ and $\\lim_{n\\rightarrow\\infty}b_n=0$.
\n\nSince this common ratio is negative, each time we multiply by it we alternate the sign of the term, that is, this series is also an alternating series. If we wanted to, we could pull the common factor of $-1$ out and write our series as a positive number multiplied by $(-1)^{n+1}$:
\n$\\displaystyle\\sum_{n=0}^\\infty (-1)^{n+1}\\var{abs(a)}\\left(\\var[fractionNumbers]{abs(r)}\\right)^{\\var{start}}\\left(\\var[fractionNumbers]{abs(r)}\\right)^{n}$
\nThe alternating series $\\sum_{n=0}^\\infty (-1)^{n+1} b_n$ where $b_n>0$, converges if $b_{n+1}\\le b_n$ for all $n$ and $\\lim_{n\\rightarrow\\infty}b_n=0$.
\nTherefore, we also know our series converges based on this alternating series test.
\nTherefore, we also know our series diverges based on this alternating series test.
\n", "name": "Series: geometric series", "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/"}]}