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Determine a finite sum and a limiting sum of a geometric series written using sigma notation. I'm not sure how to eliminate rounding errors...

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A series is given by \\[\\sum_{k=\\var{start}}^{\\var{stop}}\\var{a}\\left(\\var[fractionNumbers]{r}\\right)^k\\]

\\[\\sum_{k=\\var{start}}^{\\var{stop}}\\var{a}\\times \\var{r}^k\\]

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What is the first term in this series? [[0]]

What does this series sum to? [[0]]

Note: It is likely your calculator will give you an answer that is not accurate enough. If this is the case, you can input your expression such as $\\frac{2\\times3^4(1-1.7^{10})}{-0.7}$ using the syntax 2*3^4*(1-1.7^10)/(-0.7)

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Does the infinite series $\\sum_{k=\\var{start}}^{\\infty}\\var{a}\\left(\\var[fractionNumbers]{r}\\right)^k$ $\\sum_{k=\\var{start}}^{\\infty}\\var{a}\\times \\var{r}^k$ converge or diverge? [[0]]

What does it converge to? [[1]]

converge

", "

diverge

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$\\displaystyle\\sum_{k=\\var{start}}^{\\var{stop}}\\var{a}\\left(\\var[fractionNumbers]{r}\\right)^k$ $\\displaystyle\\sum_{k=\\var{start}}^{\\var{stop}}\\var{a}\\times \\var{r}^k$ is a nice way of writing

\\[\\var{a}\\left(\\var[fractionNumbers]{r}\\right)^\\var{start}+\\var{a}\\left(\\var[fractionNumbers]{r}\\right)^\\var{start+1}+\\var{a}\\left(\\var[fractionNumbers]{r}\\right)^\\var{start+2}+\\ldots+\\var{a}\\left(\\var[fractionNumbers]{r}\\right)^\\var{stop}\\] \\[\\var{a}\\times \\var{r}^\\var{start}+\\var{a}\\times \\var{r}^\\var{start+1}+\\var{a}\\times \\var{r}^\\var{start+2}+\\dots+\\var{a}\\times \\var{r}^\\var{stop}\\]

Therefore the first term is $\\simplify[fractionNumbers]{{a*r^start}}$.

Notice each term in the series is just the last term multiplied by $\\var[fractionNumbers]{r}$ that is, the common ratio between consecutive terms is $\\var[fractionNumbers]{r}$. Therefore, the series is a geometric series and we can determine its sum using the formula \\[S_n=\\frac{a(1-r^n)}{1-r}\\] where $a$ is the first term, $r$ is the common ratio and $n$ is the number of terms in the sum.

In our case we have $a=\\simplify[fractionNumbers]{{a*r^start}}$, $r=\\var[fractionNumbers]{{r}}$, and $n=\\var{n+1}$, so the sum is \\[\\begin{align}S_{\\var{n+1}}&=\\frac{\\simplify[fractionNumbers]{{a*r^start}}\\left(1-\\left(\\var[fractionNumbers]{{r}}\\right)^\\var{n+1}\\right)}{1-\\left(\\var[fractionNumbers]{{r}}\\right)}\\\\&=\\frac{\\simplify[fractionNumbers]{{a*r^start}}\\left(1-\\left(\\var[fractionNumbers]{{r}}\\right)^\\var{n+1}\\right)}{\\var[fractionNumbers]{1-r}}\\\\&=\\simplify[fractionNumbers,simplifyFractions]{{(a*r^start)/(1-r)}}\\left(1-\\left(\\var[fractionNumbers]{{r}}\\right)^\\var{n+1}\\right)\\end{align}\\]

Notice each term in the series is actually the same! So we just have $\\var{n+1}$ lots of $\\var{a}$ and so our sum is simply $\\var{a*(n+1)}$.

Since $r$ is negative and $n$ is odd, this is the same as $\\simplify[fractionNumbers,simplifyFractions]{{(a*r^start)/(1-r)}}\\left(1+\\left(\\var[fractionNumbers]{-r}\\right)^\\var{n+1}\\right)$$=\\var[fractionNumbers]{sum}$.

Since $r$ is negative and $n$ is even, this is the same as $\\simplify[fractionNumbers,simplifyFractions]{{(a*r^start)/(1-r)}}\\left(1-\\left(\\var[fractionNumbers]{-r}\\right)^\\var{n+1}\\right)$ $=\\var[fractionNumbers]{sum}$.

Also, notice that since $r=-1$ each term is the same as the last except the sign has been swapped! This means the result of the sum oscillates between $\\var[fractionNumbers]{a*r^start}$ and $0$. So if there are an even number of terms the sum will be $0$ but if there are an odd number of terms the sum will be $\\var[fractionNumbers]{a*r^start}$.

For this question, you should be able to write your answer as this expression instead of using the calculator and getting (possibly) a decimal approximation.

A geometric series converges if and only if $|r|<1$ (that is, $-1<r<1$). We can realise this by taking the limit as $n$ goes to infinity of the sum of the first $n$ terms. Recall the sum of the first $n$ terms is $S_n=\\frac{a(1-r^n)}{1-r}$. If $|r|<1$ and we take the limit as $n\\rightarrow \\infty$, then $r^n\\rightarrow 0$ and therefore

\\[S_\\infty=\\frac{a}{1-r}, \\qquad \\text{for }|r|<1.\\]

In our case, $|r|=\\var[fractionNumbers]{abs(r)}<1$ and so the series converges to \\[\\frac{\\var[fractionNumbers]{a*r^start}}{1-\\var[fractionNumbers]{r}}=\\simplify[fractionNumbers]{{a*r^start/(1-r)}}\\]

In our case, $|r|=\\var[fractionNumbers]{abs(r)}\\not<1$ and so the series diverges.

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