2019 SAJC P2 Q4

2019 SAJC P2 Q4

8 marks
Prelims
  1. Given that \(\mathrm{f}\left( r \right)=\frac{r}{2{{\,}^{r}}}\), by considering \(\mathrm{f}\left( r+1 \right)-\mathrm{f}\left( r \right)\), find \(\sum\limits_{r=\,1}^{n}{\frac{1-r}{2{{\,}^{r+1}}}}\).[3]
    1. Cauchy’s root test states that a series of the form \(\sum\limits_{r=0}^{\infty }{{{a}_{r}}}\) (where \({{a}_{r}}>0\) for all \(r\)) converges when \(\mathop{\lim}\limits_{n\to \infty }\,\sqrt[n]{{{a}_{n}}}<1\), and diverges when \(\mathop{\lim}\limits_{n\to \infty }\,\sqrt[n]{{{a}_{n}}}>1\). When \(\mathop{\lim}\limits_{n\to \infty }\,\sqrt[n]{{{a}_{n}}}=1\), the test is inconclusive. Using the test and given that \(\mathop{\lim}\limits_{n\to \infty }\,\sqrt[n]{{{n}^{p}}}=1\) for all positive \(p\), explain why the series \(\sum\limits_{r=0}^{\infty }{\frac{2{{\,}^{r}}{{r}^{x}}}{{{3}^{\,r}}}}\) converges for all positive values of \(x\).[3]
    2. By considering \({{\left( 1-y \right)}^{-2}}=1+2y+3{{y}^{2}}+4{{y}^{3}}+...\), evaluate \(\sum\limits_{r=0}^{\infty }{\frac{2{{\,}^{r}}{{r}^{x}}}{{{3}^{\,r}}}}\) for the case when \(x=1\).[2]

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Answer:(a) \(\dfrac{n+1}{2^{n+1}} - \dfrac{1}{2}\); (b)(ii) \(6\)

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