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10.6 Alternating Series

Series w/ alternating signs

\[ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots \]Alternating Harmonic Series

\[ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{2k-1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \dots \]Leibnitz Series

\[ = \frac{\pi}{4} \text{ (converges to } \frac{\pi}{4} \text{)} \]

\( (-1)^k \) or variations \( \rightarrow \) alternating signs

power of \( (-1) \) shifted by 2

\[ \sum_{k=1}^{\infty} (-1)^{k-1} \frac{1}{2k-1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \dots \]

shifting power by 2 while keeping the same starting k does not change the series

General form:

\[ \sum_{k=1}^{\infty} (-1)^{k+1} a_k \]

\( a_k \) is always non-negative

\[ \sum_{k=1}^{\infty} (-1)^{k+1} \left( \frac{1}{k} \right) \]

\( a_k = \frac{1}{k} \)

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Two things need to happen for an alternating series to converge

\( a_{k+1} \leq a_k \) after some \( k \)
\( \rightarrow \) magnitude does not get bigger
\( \lim_{k \to \infty} a_k = 0 \)
\( \rightarrow \) the divergence test

The Alternating Series Test

Look at the alternating Harmonic series

\[ \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots \]

clearly, \( \lim_{k \to \infty} \frac{1}{k} = 0 \) (passes the divergence test)

and \( \frac{1}{k} \) clearly decreases as \( k \) increases

\[ \frac{1}{k+1} \leq \frac{1}{k} \text{ for any } k \]

So, the Alt. Harmonic Series converges (even though the "regular" Harmonic does not)

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Alternating Harmonic Series Convergence

What does \(\sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k}\) converge to?

\[\sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \dots\]

Partial Sums

\(S_1 = 1\)

\(S_2 = 1 - \frac{1}{2} = \frac{1}{2}\)

\(S_3 = 1 - \frac{1}{2} + \frac{1}{3} = 0.8333\)

\(S_4 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = 0.5833\)

\(S_5 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} = 0.7833\)

What we add back < what we took away

Should settle down

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Partial Sums of \(\sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k}\)

The following graph visualizes the behavior of the partial sums as the number of terms increases. The values oscillate but gradually approach a specific limit.

Figure: Scatter plot showing partial sums oscillating and converging toward a horizontal line at approximately 0.7.

Observation:

Sum looks to be around 0.7

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Alternating Series Convergence Example

Example: Does \[ \sum_{k=0}^{\infty} \frac{(-1)^k}{k^6+9} \] converge?

\[ = \frac{1}{9} - \frac{1}{10} + \frac{1}{73} - \dots \]

Convergence Test

To converge:

Is \( \lim_{k \to \infty} \frac{1}{k^6+9} = 0 \)? yes
Is \( \frac{1}{k^6+9} \) nonincreasing? yes

To be sure:

\[ \frac{d}{dk} \left( \frac{1}{k^6+9} \right) \le 0 \]

\[ \frac{-6k^5}{(k^6+9)^2} < 0 \]

Numerator: \( -6k^5 \) is negative for \( k > 0 \)

Denominator: \( (k^6+9)^2 \) is positive (squared)

So \( a_k \) is always decreasing.

So, \[ \sum_{k=0}^{\infty} \frac{(-1)^k}{k^6+9} \] converges.

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Estimating Sum of an Alternating Series

If \[ \sum_{k=1}^{\infty} (-1)^{k+1} a_k \] converges, then:

Notice the true sum \( S \) is always between two consecutive partial sums.

Figure: A coordinate graph with axes k and S_k. Points S_1, S_2, S_3, etc., oscillate above and below a dashed line labeled true sum S.

\[ S_k \le S \le S_{k+1} \]

or

\[ S_{k+1} \le S \le S_k \]

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\[ S_k \le S \le S_{k+1} \]

Subtract \( S_k \)

\[ 0 \le |S - S_k| \le |S_{k+1} - S_k| \]

how far true sum is from a partial sum

how far are two partial sums apart

\( |S_{k+1} - S_k| = ? \)

\[ \sum_{k=1}^{\infty} (-1)^{k+1} a_k = a_1 - a_2 + a_3 - a_4 + a_5 - \dots \]

\( S_1 = a_1 \)
\( S_2 = a_1 - a_2 \) → \( |S_2 - S_1| = |a_2| \)
\( S_3 = a_1 - a_2 + a_3 \) → \( |S_3 - S_2| = |a_3| \)
\( \vdots \)
\( |S_{k+1} - S_k| = |a_{k+1}| \)

so, \[ 0 \le |S - S_k| \le |a_{k+1}| \]

the partial sum \( S_k \) is no more than the magnitude of the next term away from the true sum

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example

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

\[ = \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \right) + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \dots \]

\( S_4 = \frac{7}{12} \)

\( a_5 = \frac{1}{5} \)

how close is \( S_4 \) to the true sum?

\[ 0 \le |S - S_4| \le |a_5| \]

\[ 0 \le |S - \frac{7}{12}| \le \frac{1}{5} \]

so, the true sum of the series is no more than \( \frac{1}{5} \) away from \( \frac{7}{12} \)

\[ \frac{7}{12} - \frac{1}{5} \le S \le \frac{7}{12} + \frac{1}{5} \]

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If \(\sum_{k=1}^{\infty} |a_k|\) converges then we say \(\sum_{k=1}^{\infty} a_k\) converges absolutely.

For example, \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2}\) is absolutely convergent

because \(\sum_{k=1}^{\infty} \left| \frac{(-1)^{k+1}}{k^2} \right| = \sum_{k=1}^{\infty} \frac{1}{k^2}\) converges.

If \(\sum_{k=1}^{\infty} |a_k|\) diverges BUT \(\sum_{k=1}^{\infty} a_k\) converges

\(\rightarrow \sum_{k=1}^{\infty} a_k\) converges conditionally

For example, \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\) converges but

\[ \sum_{k=1}^{\infty} \left| \frac{(-1)^{k+1}}{k} \right| = \sum_{k=1}^{\infty} \frac{1}{k} \text{ diverges} \]

So, \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\) is conditionally convergent.