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Consider the geometric series $\sum_{n=0}^\infty ar^n$ where $a=1$ and $r=-\frac{1}{2}$. Since $|r| < 1$, the series converges to $S = \sum_{n=0}^\infty ar^n = \frac{1}{1+\frac{1}{2}} = \frac{2}{3}$.

I would like to arrive at the same sum by computing $\lim_{N \to \infty} S_N$ where $S_N$ is the partial sum of $N$ terms of the goemetric series.

First few terms of the geometric series are: $1, -\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}, \frac{1}{16}, -\frac{1}{32}, \cdots$.

Here are my attempts to find $S_N$: \begin{align*} S_1 &= 1 = \frac{2^{(1-1)}}{2^{(1-1)}} \\ S_2 &= 1 - \frac{1}{2} = \frac{1}{2} = \frac{2^{(2-1)} - 1}{2^{(2-1)}} \\ S_3 &= \frac{1}{2} + \frac{1}{4} = \frac{3}{4} = \frac{2^{(3-1)} - 1}{2^{(3-1)}} \\ S_4 &= \frac{3}{4} - \frac{1}{8} = \frac{5}{8} = \frac{2^{(4-1)} - 3}{2^{(4-1)}} \\ \cdots \\ S_N &= ?? \end{align*} \begin{align*} S_1 &= 1 - \frac{0}{1} \\ S_2 &= 1 - \frac{1}{2} \\ S_3 &= 1 - \frac{1}{2} + \frac{1}{4} = 1 - \frac{1}{4} \\ S_4 &= 1 - \frac{1}{4} - \frac{1}{8} = 1 - \frac{3}{8} \\ \cdots \\ S_N &= ?? \end{align*}

I cannot find a pattern that will help me find $S_N$.

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  • $\begingroup$ Do you know the formula for summing up $N$ terms of a GP? $\endgroup$ – Calvin Lin Jan 12 '14 at 23:07
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Hint:

$$a + ar + ar^2 + \ldots + ar^{n-1} = \frac{ a( 1- r^n) } { (1-r) }. $$

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    $\begingroup$ The classical formula: @Ravi to see how it's derived, take the finite series $ \ S_N \ = \ a + ar + \ldots + ar^{n-1} \ $ , multiply it through by $ \ r \ $ , then subtract $ \ rS_N \ $ from $ \ S_N \ $ . $\endgroup$ – colormegone Jan 12 '14 at 23:14
  • $\begingroup$ Realized, I don't have to find a pattern to find S_N. Derivation of S = a / (1-r) already has an expression for S_N. $\endgroup$ – Ravi Jan 12 '14 at 23:31
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Note that $$S_1=1=\color{green}{\bf\frac23}\color{red}{\bf+\frac13},\ S_2=\frac12=\color{green}{\bf\frac23}\color{red}{\bf-\frac1{2\cdot3}},\ S_3=\frac34=\color{green}{\bf\frac23}\color{red}{\bf+\frac1{4\cdot3}},\ S_4=\frac58=\color{green}{\bf\frac23}\color{red}{\bf-\frac1{8\cdot3}},\ldots$$ As computed by @Ravi in a comment below, $$ S_N=\color{green}{\bf\frac23}+\color{red}{\bf(-1)^{N+1}\frac1{3\cdot2^{N-1}}}=\color{green}{\bf\frac23}-\frac23\cdot\left(-\frac12\right)^N. $$

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  • $\begingroup$ Another expression for $S_N$ based on its converging value. $S_N = \frac{2}{3} + (-1)^{(N+1)} \frac{1} {2^{(N-1)} \times 3}$. $\endgroup$ – Ravi Jan 12 '14 at 23:49
  • $\begingroup$ Right. From which the limit (and the rate of convergence to this limit) should be obvious. $\endgroup$ – Did Jan 13 '14 at 0:03

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