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10.7 Ratio and Root Tests

Given \(\sum_{k=1}^{\infty} a_k\), if \(\sum_{k=1}^{\infty} |a_k|\) converges, then series \(\sum_{k=1}^{\infty} a_k\) converges absolutely.

e.g. \(\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2}\) converges and \(\sum_{k=1}^{\infty} \frac{1}{k^2}\) converges

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

if a series converges absolutely, then it converges

if \(\sum_{k=1}^{\infty} a_k\) converges, but \(\sum_{k=1}^{\infty} |a_k|\) does not, then \(\sum_{k=1}^{\infty} a_k\) converges conditionally.

e.g. \(\sum_{k=1}^{\infty} \frac{(-1)}{k}\) converges but \(\sum_{k=1}^{\infty} \frac{1}{k}\) diverges

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Ratio and Root Tests tell us if a series converges absolutely (and therefore converges)

Ratio Test

Given \(\sum_{k=1}^{\infty} a_k\), the series converges absolutely (and therefore converges)

if \[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1 \]

series diverges if \[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| > 1 \]

test is inconclusive if \[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = 1 \]

why?

if \[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = r \text{ , then this says if } k \text{ is large,} \]

\[ |a_{k+1}| \approx r |a_k| \rightarrow \text{like a geometric series which converges if } |r| < 1 \]

this is why we need \[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1 \]

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Ratio Test Example

example

\[ \sum_{k=1}^{\infty} k \left( \frac{1}{4} \right)^k \]

Let \( a_k = k \left( \frac{1}{4} \right)^k \)

\[ a_k = k \left( \frac{1}{4} \right)^k \]

\[ a_{k+1} = (k+1) \left( \frac{1}{4} \right)^{k+1} \]

\[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} \left| \frac{(k+1) \left( \frac{1}{4} \right)^{k+1}}{k \left( \frac{1}{4} \right)^k} \right| \]

\[ = \lim_{k \to \infty} \left| \frac{k+1}{k} \cdot \left( \frac{1}{4} \right) \right| = 1 \cdot \left| \frac{1}{4} \right| = \frac{1}{4} < 1 \]

so this series converges (absolutely)

means the tail of this series \( (k \to \infty) \) behaves like a geo. series w/ \( r = \frac{1}{4} \)

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Ratio Test with Factorials

ratio test handles factorials well

example

\[ \sum_{k=1}^{\infty} \frac{k!}{(2k+6)!} \]

Let \( a_k = \frac{k!}{(2k+6)!} \)

\[ a_{k+1} = \frac{(k+1)!}{(2(k+1)+6)!} = \frac{(k+1)!}{(2k+8)!} \]

\[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} \left| \frac{\frac{(k+1)!}{(2k+8)!}}{\frac{k!}{(2k+6)!}} \right| = \lim_{k \to \infty} \left| \frac{(k+1)!}{(2k+8)!} \cdot \frac{(2k+6)!}{k!} \right| \]

\[ = \lim_{k \to \infty} \left| \frac{(k+1)!}{k!} \cdot \frac{(2k+6)!}{(2k+8)!} \right| \]

\[ 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \]\[ k! = k \cdot (k-1) \cdot (k-2) \cdots (1) \]

\[ = \lim_{k \to \infty} \left| \frac{(k+1)(k)(k-1)(k-2) \cdots (1)}{(k)(k-1)(k-2) \cdots (1)} \cdot \frac{(2k+6)(2k+5) \cdots (1)}{(2k+8)(2k+7)(2k+6)(2k+5) \cdots (1)} \right| \]

\[ = \lim_{k \to \infty} \left| \frac{k+1}{(2k+8)(2k+7)} \right| = 0 < 1 \]

so series converges

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Example: Convergence of a Power Series

For what values of \( x \) does

\[ \sum_{k=1}^{\infty} \frac{5x^k}{4k} \text{ converge?} \]

\( = \frac{5}{4}x + \frac{5}{8}x^2 + \frac{5}{12}x^3 + \dots \)

Ratio Test

\( a_k = \frac{5x^k}{4k} \)

\( a_{k+1} = \frac{5x^{k+1}}{4(k+1)} \)

\[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} \left| \frac{\frac{5x^{k+1}}{4(k+1)}}{\frac{5x^k}{4k}} \right| = \lim_{k \to \infty} \left| \frac{5x^{k+1}}{4k+4} \cdot \frac{4k}{5x^k} \right| \]

\[ = \lim_{k \to \infty} \left| \frac{4k}{4k+4} \cdot \frac{5x^{k+1}}{5x^k} \right| = \lim_{k \to \infty} \left| \frac{k}{k+1} \cdot x \right| = |x| \]

we want \( < 1 \)

so, \( |x| < 1 \rightarrow -1 < x < 1 \)

but notice at \( x = -1 \) and \( x = 1 \), the ratio is 1

\( \rightarrow \) Inconclusive

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we need to test \( x = -1 \) and \( x = 1 \) separately

\[ \sum_{k=1}^{\infty} \frac{5x^k}{4k} \]

when \( x = -1 \),

\( \sum_{k=1}^{\infty} \frac{5}{4} \frac{(-1)^k}{k} \rightarrow \text{converges} \)

when \( x = 1 \),

\( \sum_{k=1}^{\infty} \frac{5}{4k} = \sum_{k=1}^{\infty} \frac{5}{4} \cdot \frac{1}{k} \rightarrow \text{diverges (harmonic or p-series w/ p=1)} \)

so series converges on \( -1 \le x < 1 \)

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Root Test

Given \[ \sum_{k=1}^{\infty} a_k \] , it converges absolutely if

\[ \lim_{k \to \infty} \sqrt[k]{|a_k|} < 1 \]

Series diverges if \[ \lim_{k \to \infty} \sqrt[k]{|a_k|} > 1 \]
Inconclusive if \[ \lim_{k \to \infty} \sqrt[k]{|a_k|} = 1 \]

Why?

If \[ \lim_{k \to \infty} \sqrt[k]{|a_k|} = r \]

then for large \( k \), \[ \sqrt[k]{|a_k|} \approx r \quad \text{so} \quad |a_k| \approx r^k \]

so like a geo. series w/ ratio \( r \)

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Root test is good when \( a_k \) has some kind of \( k \) in exponent

Example

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

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

Root Test

\[ \lim_{k \to \infty} \sqrt[k]{|a_k|} = \lim_{k \to \infty} \sqrt[k]{\left( \frac{k+1}{k^2} \right)^k} = \lim_{k \to \infty} \frac{k+1}{k^2} = 0 < 1 \]

so converges

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Ratio Test Applications

Ratio Test is good with lots of series, so often it is a good first test to try.

When to use the Ratio Test:

Good with factorials
Things you don't want to integrate
Things you don't want to compare to other things

When not to use:

Not good with things that look like p-series.

Example

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

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

\[ = \lim_{k \to \infty} \left| \frac{k+1}{k} \cdot \frac{k^2+1}{(k+1)^2+1} \right| = 1 \quad \text{inconclusive} \]

Note: In the limit above, both terms \( \frac{k+1}{k} \) and \( \frac{k^2+1}{(k+1)^2+1} \) approach 1 as \( k \to \infty \).

But a comparison tells us the result quickly:

\[ \frac{k}{k^2+1} \approx \frac{k}{k^2} \approx \frac{1}{k} \text{ as } k \to \infty, \text{ so } \sum_{k=1}^{\infty} \frac{k}{k^2+1} \text{ is like } \sum \frac{1}{k} \text{ (diverges)} \]