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10.8 Choosing a Convergence Test

Summary of Tests

Divergence Test:

\[ \lim_{k \to \infty} a_k = 0 \rightarrow \text{series might converge, test more} \]\[ \lim_{k \to \infty} a_k \neq 0 \rightarrow \text{series diverges} \]

Integral Test:

\[ \sum_{k=1}^{\infty} a_k \text{ converges if } \int_{1}^{\infty} a_k(x) dx \text{ converges} \]

p-series Test:

\[ \sum_{k=1}^{\infty} \frac{1}{k^p} \text{ converges if } p > 1 \]

Geometric Series:

\[ \sum_{k=0}^{\infty} ar^k \text{ converges if } |r| < 1 \]\[ \text{Sum} = \frac{a}{1-r} \quad (a = \text{first term}) \]

Direct Comparison Test:

Often compare to p-series or geometric

If \( a_k \leq \text{terms of convergent series} \) then \( \sum a_k \) converges

If \( a_k \geq \text{terms of divergent series} \) then \( \sum a_k \) diverges

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Limit Comparison Test:

\( \sum a_k \) unknown, \( \sum b_k \) known (typically a p-series or geo. series)

\[ \lim_{k \to \infty} \frac{a_k}{b_k} = c \]\[ \text{if } 0 < c < \infty \text{ then } \sum a_k \text{ and } \sum b_k \text{ BOTH converge or BOTH diverge} \]

Alternating Series Test:

\[ \sum_{k=1}^{\infty} (-1)^{k+1} a_k \text{ converges if } \lim_{k \to \infty} a_k = 0 \]\[ \text{AND } a_k \text{ is eventually nonincreasing} \]

Ratio Test:

\[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1 \quad \text{converges} \]\[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| > 1 \quad \text{diverges} \]\[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = 1 \text{ or DNE } \rightarrow \text{inconclusive, test more (mistake in lecture + video)} \]

Root Test:

\[ \lim_{k \to \infty} \sqrt[k]{|a_k|} < 1 \quad \text{converges} \]\[ \lim_{k \to \infty} \sqrt[k]{|a_k|} > 1 \quad \text{diverges} \]\[ \lim_{k \to \infty} \sqrt[k]{|a_k|} = 1 \text{ or DNE } \rightarrow \text{inconclusive (mistake in lecture + video)} \]

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Series Convergence Examples

Example 1

\[ \sum_{k=1}^{\infty} \frac{11k^5 - 9k^3 + 5k + 10}{12k^5 + k^2 - k + 110} \]

Always do the Div Test first.

\[ \lim_{k \to \infty} a_k \neq 0 \quad \text{diverges} \]

Example 2

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

fails divergence test:

\[ \lim_{k \to \infty} \left( 1 - \frac{1}{k} \right)^k = e^{-1} \neq 0 \]

diverges

Example 3

\[ \sum_{k=1}^{\infty} \frac{1 + \sin 9k}{k^2} \]

passes div. test? yes, test more

Comparison is good here, because the terms "look like" \( \frac{1}{k^2} \)

\( -1 \leq \sin 9k \leq 1 \)

\( 0 \leq 1 + \sin 9k \leq 2 \)

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\[ 0 \leq \frac{1 + \sin 9k}{k^2} \leq \frac{2}{k^2} \]

so the terms of this series are less than or equal to those of a convergent series \( \left( \sum \frac{2}{k^2} \right) \)

therefore

\[ \sum_{k=1}^{\infty} \frac{1 + \sin 9k}{k^2} \leq \sum_{k=1}^{\infty} \frac{2}{k^2} = C \text{ (convergent)} \]

therefore

\[ \sum_{k=1}^{\infty} \frac{1 + \sin 9k}{k^2} \leq C \text{ so converges} \]

Try limit comparison

compare to \( \sum_{k=1}^{\infty} \frac{1}{k^2} = \sum_{k=1}^{\infty} b_k \)

\[ \lim_{k \to \infty} \left\{ \frac{a_k}{b_k} \right\} = \lim_{k \to \infty} \frac{\frac{1 + \sin 9k}{k^2}}{\frac{1}{k^2}} = \lim_{k \to \infty} 1 + \sin 9k \quad \text{DNE} \]

this is not easy to conclude

Direct Comp. is better here

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Example: Series Convergence

\[ \sum_{k=2}^{\infty} \frac{5}{k(\ln k)^9} \]

passes div. test? yes, test more

integral test is good because \( \frac{5}{x(\ln x)^9} \) can be integrated with \( u = \ln x \), \( du = \frac{1}{x} \)

\[ \int_{2}^{\infty} \frac{5}{x(\ln x)^9} dx \text{ converges?} \]

\[ = \lim_{b \to \infty} \int_{2}^{b} \frac{5}{x(\ln x)^9} dx \]

\( u = \ln x \)
\( du = \frac{1}{x} dx \)

\[ = \lim_{b \to \infty} \int_{\ln 2}^{\ln b} \frac{5}{u^9} du \neq \infty \]

converges since \( 9 > 1 \)

so \( \sum_{k=2}^{\infty} \frac{5}{k(\ln k)^9} \) converges

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Comparison?

maybe to \( \frac{5}{k^8} \)?

\[ \frac{5}{k(\ln k)^9} \quad \quad \ln k < k \]

so

\[ \frac{5}{k(\ln k)^9} \geq \frac{5}{k^8} \]

\( k(\ln k)^9 \) is smaller than \( k \cdot k^7 = k^8 \)

so direct comp. does not work, but a limit comp to \( \sum \frac{5}{k^8} \) might work

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Example: Series Convergence

\[ \sum_{k=1}^{\infty} \left( \sqrt{16k^4 + 1} - 4k^2 \right) \]

Divergence Test

We check the limit of the terms as \( k \to \infty \):

\[ \lim_{k \to \infty} \left( \sqrt{16k^4 + 1} - 4k^2 \right) = 0? \]

Note that as \( k \to \infty \), both terms approach infinity, leading to an indeterminate form of the type \( \infty - \infty = ? \)

We can't conclude at least in its current form.

Rationalizing the Expression

To evaluate the limit and the series, we multiply by the conjugate:

\[ \frac{\sqrt{16k^4 + 1} - 4k^2}{1} \cdot \frac{\sqrt{16k^4 + 1} + 4k^2}{\sqrt{16k^4 + 1} + 4k^2} = \frac{16k^4 + 1 - 16k^4}{\sqrt{16k^4 + 1} + 4k^2} \]

\[ = \frac{1}{\sqrt{16k^4 + 1} + 4k^2} \]

The series can now be rewritten as:

\[ \sum_{k=1}^{\infty} \frac{1}{\sqrt{16k^4 + 1} + 4k^2} \]

passes div. test

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Asymptotic Behavior

When \( k \) is large, the term behaves as follows:

\[ \frac{1}{\sqrt{16k^4 + 1} + 4k^2} \text{ looks like } \frac{1}{\sqrt{16k^4} + 4k^2} = \frac{1}{8k^2} \]

So comparison or limit comparison should work well.