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You are given the series

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\$\\sum_{k=\\var{start}}^\\infty\\var{a}\\left(\\var[fractionNumbers]{r}\\right)^k.\$

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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This is a [] [].

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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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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 $|r|\\ge1$.

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Our 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:

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$\\displaystyle\\sum_{k=\\var{start}}^\\infty \\var{a}\\left(\\var[fractionNumbers]{r}\\right)^k=\\sum_{k=\\var{start}}^\\infty \\var{a}\\left(\\var[fractionNumbers]{r}\\right)^{\\var{start}}\\left(\\var[fractionNumbers]{r}\\right)^{k-\\var{start}}$

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we now make the substitution $n=k-\\var{start}$ (which means when $k=\\var{start}$, $n=0$) and we have

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$\\displaystyle\\sum_{n=0}^\\infty \\var{a}\\left(\\var[fractionNumbers]{r}\\right)^{\\var{start}}\\left(\\var[fractionNumbers]{r}\\right)^{n}$

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\n

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and 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.

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So we have a convergent divergent geometric series with common ratio $r=\\var[fractionNumbers]{r}$.

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Since 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 $(-1)^n$:

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$\\displaystyle\\sum_{n=0}^\\infty (-1)^n\\var{abs(a)}\\left(\\var[fractionNumbers]{abs(r)}\\right)^{\\var{start}}\\left(\\var[fractionNumbers]{abs(r)}\\right)^{n}$

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The 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 $\\lim_{n\\rightarrow\\infty}b_n=0$.

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Since 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 $(-1)^{n+1}$:

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$\\displaystyle\\sum_{n=0}^\\infty (-1)^{n+1}\\var{abs(a)}\\left(\\var[fractionNumbers]{abs(r)}\\right)^{\\var{start}}\\left(\\var[fractionNumbers]{abs(r)}\\right)^{n}$

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The 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 $\\lim_{n\\rightarrow\\infty}b_n=0$.

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Therefore, we also know our series converges based on this alternating series test.

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Therefore, we also know our series diverges based on this alternating series test.

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