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10.3 Infinite Series

NOT on exam 2

\[ \sum_{k=1}^{\infty} a_k = a_1 + a_2 + a_3 + \dots \]

\( a_k \) : \( k^{\text{th}} \) term, usually formula is given

many kinds of series to study

today: geometric series

telescoping series

geometric series: common ratio between the terms

for example, \( \frac{1}{4} + \frac{1}{12} + \frac{1}{36} + \frac{1}{108} + \dots \) ratio: \( r = \frac{1}{3} \)

we can write a geometric series in this form:

\[ a + ar + ar^2 + ar^3 + ar^4 + \dots = \sum_{k=0}^{\infty} ar^k \]

\( \frac{1}{4} + \frac{1}{12} + \frac{1}{36} + \frac{1}{108} + \dots = \frac{1}{4} (1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots) \)

\( = \frac{1}{4} + \frac{1}{4}(\frac{1}{3}) + \frac{1}{4}(\frac{1}{3})^2 + \frac{1}{4}(\frac{1}{3})^3 + \dots \)

\[ = \sum_{k=0}^{\infty} \frac{1}{4} (\frac{1}{3})^k \]

Note: In the expansion above, the first term is \( a \), and subsequent terms are multiplied by the ratio \( r \).

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the starting \( k \) value doesn't really matter

\[ \sum_{k=0}^{\infty} ar^k = a + ar + ar^2 + ar^3 + \dots \]

notice \( \sum_{k=1}^{\infty} ar^{k-1} \) is the same series

\( = \underset{k=1}{ar^{1-1}} + \underset{k=2}{ar^{2-1}} + \underset{k=3}{ar^{3-1}} + \dots \)

\( = ar^0 + ar + ar^2 + \dots = a + ar + ar^2 + ar^3 + \dots \)

The two expressions above represent the same series.

the partial sums of a geometric series

\( \frac{1}{4} + \frac{1}{12} + \frac{1}{36} + \frac{1}{108} + \frac{1}{324} + \frac{1}{972} + \dots \)

1st partial sum \( S_1 = \frac{1}{4} \)

\( S_2 = \frac{1}{4} + \frac{1}{12} \)

\( S_3 = \frac{1}{4} + \frac{1}{12} + \frac{1}{36} \)

\( \vdots \)

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\[ \sum_{k=0}^{\infty} ar^k = a + ar + ar^2 + ar^3 + \dots \]

\( S_1 = a = ar^0 \)

\( S_2 = a + ar^1 \)

\( S_3 = a + ar + ar^2 \)

\( S_4 = a + ar + ar^2 + ar^3 \)

pattern: \( n^{\text{th}} \) partial sum ends w/ \( n-1 \) power of \( r \)

\( \vdots \)

\[ S_n = a + ar + ar^2 + \dots + ar^{n-1} \quad \text{--- (1)} \]

let's find a formula for \( S_n \) w/o adding \( n \) terms

multiply by \( r \)

\[ rS_n = ar + ar^2 + ar^3 + \dots + ar^n \quad \text{--- (2)} \]

\( (1) - (2) \)

\( S_n - rS_n = a - ar^n \)

\( S_n(1 - r) = a - ar^n \)

\( \rightarrow \)
\[ S_n = \frac{a - ar^n}{1 - r} \]
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\[ \frac{1}{4} + \frac{1}{36} + \frac{1}{108} + \dots = \sum_{k=0}^{\infty} \frac{1}{4} \left( \frac{1}{3} \right)^k \rightarrow a = \frac{1}{4}, \, r = \frac{1}{3} \]

\( 7^{\text{th}} \) partial sum (w/o adding 7 terms)

\[ S_7 = \frac{a - ar^7}{1 - r} = \frac{\frac{1}{4} - \frac{1}{4} \left( \frac{1}{3} \right)^7}{1 - \frac{1}{3}} = \dots = \boxed{\frac{1093}{2916}} \]

bigger question: does \( S_n \rightarrow \text{finite value as } n \rightarrow \infty \)?

in other words, does \( \sum_{k=0}^{\infty} ar^k \) converge? \( \lim_{n \rightarrow \infty} S_n = L \)?

\[ \lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} \frac{a(1 - r^n)}{1 - r} = \frac{a}{1 - r} \lim_{n \rightarrow \infty} (1 - r^n) \]

limit exists only if \( r^n \) as \( n \rightarrow \infty \) exists

and that happens if \( |r| < 1 \) or \( -1 < r < 1 \)

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Geometric Series Convergence

So, for \[ \sum_{k=0}^{\infty} ar^k \] it converges if \( |r| < 1 \).

And its sum is

\[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{a(1-r^n)}{1-r} = \frac{a}{1-r} \]

Note on the sum formula \( \frac{a}{1-r} \):

  • a: first term
  • r: common ratio
  • The term \( r^n \) goes to 0 as \( n \to \infty \).

Examples

Try on: \[ \frac{1}{4} + \frac{1}{12} + \frac{1}{36} + \frac{1}{108} + \dots = \sum_{k=0}^{\infty} \frac{1}{4} \left( \frac{1}{3} \right)^k \] In this case, \( a = \frac{1}{4} \) and \( r = \frac{1}{3} \).

\[ = \frac{a}{1-r} = \frac{\frac{1}{4}}{1 - \frac{1}{3}} = \frac{\frac{1}{4}}{\frac{2}{3}} = \frac{3}{8} \]

\[ 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \dots = \sum_{k=0}^{\infty} \left( -\frac{1}{2} \right)^k \] In this case, \( a = 1 \) and \( r = -\frac{1}{2} \).

\[ = \frac{a}{1-r} = \frac{1}{1 - (-\frac{1}{2})} = \frac{2}{3} \]
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Telescoping Series

Example

\[ \sum_{k=1}^{\infty} \frac{1}{(k+11)(k+12)} \]

\[ = \underbrace{\frac{1}{12 \cdot 13}}_{k=1} + \underbrace{\frac{1}{13 \cdot 14}}_{k=2} + \frac{1}{14 \cdot 15} + \frac{1}{15 \cdot 16} + \dots = ? \] Not clear where this is going.

Partial Fraction Decomposition

\[ \frac{1}{(k+11)(k+12)} = \frac{A}{k+11} + \frac{B}{k+12} = \frac{1}{k+11} - \frac{1}{k+12} \]

Rewriting the series: \[ \sum_{k=1}^{\infty} \left( \frac{1}{k+11} - \frac{1}{k+12} \right) \] \[ = \underbrace{\left( \frac{1}{12} - \frac{1}{13} \right)}_{k=1} + \underbrace{\left( \frac{1}{13} - \frac{1}{14} \right)}_{k=2} + \left( \frac{1}{14} - \frac{1}{15} \right) + \dots \]

Only first and last terms survive in the partial sums:

  • \( S_3 = \frac{1}{12} - \frac{1}{15} \)
  • \( S_4 = \frac{1}{12} - \frac{1}{16} \)
  • \( S_5 = \frac{1}{12} - \frac{1}{17} \)
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Generalizing for Partial Sums

now generalize for \(S_n\)

\[S_n = \frac{1}{12} - \frac{1}{12+n}\]

does this series converge? yes, because \(\lim_{n \to \infty} S_n\) exists

what does it converge to? \(\frac{1}{12}\)

if we keep adding terms, the sum approaches \(\frac{1}{12}\)

The Starting Index and Convergence

the starting \(k\) does NOT affect convergence

if a series converges, changing starting \(k\) results in a convergent series

for example, \(\sum_{k=0}^{\infty} (\frac{1}{2})^k\) converges because \(|r| < 1\)

\[= \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \right) + \left( \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \dots \right) = 2 = \frac{1}{1-\frac{1}{2}} = 2\]

\(\sum_{k=0}^{3} (\frac{1}{2})^k\)

finite series

ALWAYS converges

\(\sum_{k=4}^{\infty} (\frac{1}{2})^k\)

must also converge because it's a part of a convergent series

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EXAM REVIEW

MON 10/16

5:30 - 7:30 pm

University

Church 114

A small circular inset photo of a sea otter floating in water.
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