Geometric Series

\begin{gather*} S_n = \sum_{k=0}^{n} {x^k} = 1 + x + x^2 + ... + x^n \\ (1-x)S_n = S_n - x S_n = 1 + x + x^2 + ... + x^n - (x + x^2 + ... + x^n + x^{n+1}) = 1 - x^{n+1} \\ \Rightarrow S_n = \frac{1-x^{n+1}}{1-x} \end{gather*}

and so as $n \rightarrow \infty \text{,}$

\begin{gather*} S_n \rightarrow S = \frac{1}{1-x} \end{gather*}

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The interactive activity below shows how well the partial sums approximate

$\frac{1}{1-x}$ as the number of terms increases.

xxxxxxxxxx
 
var('x,n,k')
f = 1/(1-x)
@interact
def _(n = slider(2,20,1,2)):
    Sn = sum(x^k,k,0,n)
    pretty_print(html('$S_n(x) = %s$'%str(latex(Sn))))
    G = plot(f,x,-1,0.9,color='black')
    G += plot(Sn,x,-1,0.9,color='blue')
    G += plot(abs(f-Sn),x,-1,0.9,color='red')
    T = "Partial Sums (blue) vs Infinite Series (black) and Error (red)"
    G.show(title=T,figsize=(5,4))

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Subsection 12.1.2 Alternate Forms for the Geometric Series

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Theorem 12.1.2. Generalized Geometric Series.

For

$k \in \mathbb{N}, \sum_{k=M}^{\infty} {x^k} = \frac{x^M}{1-x}$

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Proof.

\begin{align*} \sum_{k=M}^{\infty} {x^k} & = x^M \sum_{k=0}^{\infty} {x^k}\\ & = x^M \frac{1}{1-x}\\ & = \frac{x^M}{1-x} \end{align*}

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Example 12.1.3. Integrating and Differentiating to get new Power Series.

The geometric power series is a nice function which is relatively easily differentiated and integrated. In doing so, one can obtain new power series which might also be very useful in their own right. Here we develop a few which are of special interest.

Let $f(x) = \sum_{k=0}^\infty x^k = \frac{1}{1-x}\text{.}$ Then,

$\begin{gather*} f'(x) = \sum_{k=1}^{\infty} {kx^{k-1}} = \frac{1}{(1-x)^2}\\ f''(x) = \sum_{k=2}^{\infty} {k(k-1)x^{k-1}} = \frac{2}{(1-x)^3}\\ f^{(n)}(x) = \sum_{k=n}^{\infty} {k(k-1)...(k-n+1)x^{k-n}} = \frac{n!}{(1-x)^{n+1}}\\ \int f(x) dx = \sum_{k=0}^{\infty} {\frac{x^{k+1}}{k+1}} = -ln(1-x) \end{gather*}$

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Example 12.1.4. Playing with the base.

$\begin{align*} \sum_{k=0}^{\infty} {a^k x^k} & = \sum_{k=0}^{\infty} {(ax)^k}\\ & = \frac{1}{1-ax}, |x| \lt \frac{1}{a} \end{align*}$

or perhaps

$\begin{gather*} \sum_{k=0}^{\infty} {(x-b)^k} = \frac{1}{1-(x-b)}, |x-b| \lt 1 \end{gather*}$

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Example 12.1.5. Application: Converting repeating decimals to fractional form.

Consider this example:

$\begin{align*} 2.48484848... & = 2 + 0.48 + 0.0048 + 0.000048 + ...\\ & = 2 + 0.48(1 + 0.01 + 0.0001 + ... ) = 2 + 0.48 \sum_{k=0}^\infty (0.01)^k \end{align*}$

Therefore, applying the Geometric Series

$\begin{align*} 2.48484848... & = 2 + 0.48 \frac{1}{1-0.01} \\ & = 2 + 0.48 \frac{100}{99} = 2 + \frac{48}{99} \end{align*}$

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Example 12.1.6. Playing around with repeating decimals.

Certainly most students would agree that $0.333333... = \frac{1}{3} \text{.}$ So, what about $0.999999...\text{?}$ Simply follow the pattern above

$\begin{align*} 0.999999... & = 0.9 + 0.09 + 0.009 + 0.0009 + ... = 0.9(1 + 0.1 + 0.1^2 + 0.1^3 + ...\\ & = 0.9 \frac{1}{1-0.1} = 0.9 \frac{1}{0.9} = 1 \end{align*}$