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10.5 Comparison Tests

Last time:

Divergence Test: \(\sum_{k=1}^{\infty} a_k\) diverges if \(\lim_{k \to \infty} a_k \neq 0\)

but, just because \(\lim_{k \to \infty} a_k = 0\) it does NOT necessarily mean \(\sum_{k=1}^{\infty} a_k\) converges.

Integral Test: \(\sum_{k=1}^{\infty} a_k\) converges if \(\int_{1}^{\infty} a(x) dx\) converges

\(a(x)\): function based on \(a_k\)

p-series Test: \(\sum_{k=1}^{\infty} \frac{1}{k^p}\) converges if \(p > 1\)

Comparison: compare an unknown series to one that we know

\(1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \dots \to \sum\)

\(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots = \sum_{k=0}^{\infty} (\frac{1}{2})^k\) converges

geo. series \(r = \frac{1}{2} < 1\)

what about

\(\frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{9} + \frac{1}{17} + \dots = ?\)

this is not a geo series or a p-series

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notice

\(\frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{9} + \dots \leq 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = 2\) \(\left( \text{sum} = \frac{a}{1-r} = \frac{1}{1-1/2} = 2 \right)\)

because each term on the series on the left \(\leq\) that of the right series

so, \(\frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{9} + \dots \leq 2 \to\) so this series must also converge

example

\(\sum_{k=1}^{\infty} \frac{1}{k^3 + 5}\)

always check if \(\lim_{k \to \infty} a_k = 0\) (Div. Test)
does it pass the Div. Test? Yes, test more.
compare to what?
\(\to\) think about what the series looks like if \(k\) is large

\(\frac{1}{k^3 + 5}\) as \(k \to \infty\) \(\frac{1}{k^3 + 5} \approx \frac{1}{k^3}\) so, compare to \(\sum \frac{1}{k^3}\)

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does \( \sum_{k=1}^{\infty} \frac{1}{k^3} \) converge? (p-series, \( p = 3 > 1 \))

\[ \frac{1}{k^3 + 5} \leq \frac{1}{k^3} \quad \text{for } k = 1, 2, 3, \dots \]

so, \( \sum_{k=1}^{\infty} \frac{1}{k^3 + 5} \leq \sum_{k=1}^{\infty} \frac{1}{k^3} = S \) (it has a finite sum because it converges)

therefore, \( \sum_{k=1}^{\infty} \frac{1}{k^3 + 5} \leq S \) therefore converges.

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

passes the Div. Test since \( \lim_{k \to \infty} \frac{k+1}{k^2} = 0 \)

compare to what happens when \( k \) is large

\[ \frac{k+1}{k^2} \approx \frac{k}{k^2} = \frac{1}{k} \quad \text{so compare to } \sum_{k=1}^{\infty} \frac{1}{k} \text{ (divergent)} \]

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for \( k = 1, 2, 3, 4, 5, \dots \)

\[ \frac{k+1}{k^2} \geq \frac{1}{k} \quad \text{because } \frac{1}{k} = \frac{k}{k^2} \]

therefore, \( \sum_{k=1}^{\infty} \frac{k+1}{k^2} \geq \sum_{k=1}^{\infty} \frac{1}{k} = \infty \) (diverges)

\( \sum_{k=1}^{\infty} \frac{k+1}{k^2} = \infty \) (bigger \( \infty \)) so diverges

what about \( \sum_{k=1}^{\infty} \frac{k-1}{k^2} \) ?

\[ \frac{k-1}{k^2} \leq \frac{1}{k} \]

so, \( \sum_{k=1}^{\infty} \frac{k-1}{k^2} \leq \sum_{k=1}^{\infty} \frac{1}{k} = \infty \)

\( \sum_{k=1}^{\infty} \frac{k-1}{k^2} = \infty \) (a smaller \( \infty \) OR a finite number)

this comparison does NOT give a conclusive answer to convergence question

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Comparison Test for Series

If the terms of the unknown series are \(\le\) those of a convergent series, then the unknown series converges.

If the terms of the unknown series are \(\ge\) those of a divergent series \(\rightarrow\) diverges.

If the terms of the unknown series are \(\le\) those of a divergent series \(\rightarrow\) ? inconclusive (choose a different test).

(same if terms \(\ge\) convergent series)

assumption: terms are positive

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A variation is the Limit Comparison Test

\(\sum_{k=1}^{\infty} b_k\) is a known series

\(\sum_{k=1}^{\infty} a_k\) is the unknown series

If \(\lim_{k \to \infty} \frac{a_k}{b_k} = c\) (\(c > 0\) but finite)

then BOTH \(\sum_{k=1}^{\infty} a_k\) AND \(\sum_{k=1}^{\infty} b_k\) converge or BOTH diverge

because this means as \(k \to \infty\),

\(a_k \approx c b_k\)

so \(\sum a_k \approx \sum c b_k\)

in other words, the tails look alike.

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Example: Series Convergence

\[ \sum_{k=1}^{\infty} \left( \frac{k}{2k+3} \right)^k \]

Does it pass the Div. Test?

\[ \lim_{k \to \infty} \left( \frac{k}{2k+3} \right)^k = 0 \, ? \]

As \( k \to \infty \), \( \frac{k}{2k+3} \approx \frac{k}{2k} = \frac{1}{2} \)

So as \( k \to \infty \), \( \left( \frac{k}{2k+3} \right)^k \approx \left( \frac{1}{2} \right)^k \to 0 \)

This is suggesting we compare to \( \sum_{k=1}^{\infty} \left( \frac{1}{2} \right)^k \to \text{converges} \)

Let

\[ \sum_{k=1}^{\infty} b_k = \sum_{k=1}^{\infty} \left( \frac{1}{2} \right)^k \]

\[ \sum_{k=1}^{\infty} a_k = \sum_{k=1}^{\infty} \left( \frac{k}{2k+3} \right)^k \]

\[ \lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{\left( \frac{k}{2k+3} \right)^k}{\left( \frac{1}{2} \right)^k} = \lim_{k \to \infty} \left( \frac{\frac{k}{2k+3}}{\frac{1}{2}} \right)^k \]

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\[ = \lim_{k \to \infty} \left( \frac{k}{2k+3} \cdot \frac{2}{1} \right)^k = \lim_{k \to \infty} \left( \frac{2k}{2k+3} \right)^k = \frac{1}{e^{3/2}} > 0 \text{ (and not } \infty) \]

So, BOTH \( \sum_{k=1}^{\infty} \left( \frac{1}{2} \right)^k \) AND \( \sum_{k=1}^{\infty} \left( \frac{k}{2k+3} \right)^k \) converge or diverge.

Since \( \sum_{k=1}^{\infty} \left( \frac{1}{2} \right)^k \) converges (geo. series \( r = \frac{1}{2} \))

\( \sum_{k=1}^{\infty} \left( \frac{k}{2k+3} \right)^k \) converges too.