ConvergenceTests < MSC

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---+ Convergence Tests

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---++ Divergence Test

---++ P-Series
   $ Statement: <latex>1+\frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \frac{1}{5^p} + ...</latex>%BR%
 <latex>\left \{ \begin{array} {l} \mbox{ \small Converges for $p > 1$ } \\ \mbox{ \small Diverges for $p \le 1$ } \end{array} \right.</latex>

   $ Comment: 

---++ Geometric Series and related tests.
   $ Statement: <latex>1 + x + x^2 + x^3 + ...</latex>%BR%
 <latex>\left \{ \begin{array}{l} \mbox{ \small Converges to $\frac{1}{1-x}$ if $|x| < 1$ } \\ \mbox{ \small Diverges if $|x| = 1$ } \\ \mbox{ \small Diverges if $|x| > 1$ } \end{array} \right.</latex>

   $ Comment: This is the granddaddy of many series which are easy to sum. It also is the foundation for several other tests when you observe:
      * The ratio of successive terms of the _Geometric Series_ is x -- hence the _Ratio Test_
      * The ratio of the absolute values of successive terms of the _Geometric Series_ is |x| -- hence the _Ratio Test for Absolute Convergence_
      * The <i>n<sup>th</sup></i> root of the <i>n<sup>th</sup></i> term of the _Geometric Series_ is x -- hence the _Root Test_

---+++ Ratio Test
   $ Statement: <latex>\mbox{ \small Let $\sum u_k$}</latex>%BR%
 be a series with positive terms and%BR%
 <latex>\mbox{ \small $\lim_{k \rightarrow \infty} \frac{u_{k+1}}{u_k} = \rho$ }</latex>%BR%
 then:%BR%
 <latex>\left \{ \begin{array} {l} \mbox{ \small Converges if $\rho < 1$ } \\ \mbox{ \small No conclusion if $\rho = 1$ } \\ \mbox{ \small Diverges if $\rho > 1$ } \end{array} \right.</latex>

   $ Comment: Try this test when <i>u<sub>k</sub></i> involves factorials or <i>k<sup>th</sup></i> powers.

---+++ Ratio Test for Absolute Convergence
   $ Statement: <latex>\mbox{ \small Let $\sum u_k$}</latex>%BR%
 be a series with non-zero terms and%BR%
 <latex>\mbox{ \small $\lim_{k \rightarrow \infty} \frac{|u_{k+1}|}{|u_k|} = \rho$ }</latex>%BR%
 then:%BR%
 <latex>\left \{ \begin{array} {l} \mbox{ \small Converges if $\rho < 1$ } \\ \mbox{ \small No conclusion if $\rho = 1$ } \\ \mbox{ \small Diverges if $\rho > 1$ } \end{array} \right.</latex>

   $ Comment: The series need not have positive terms and need not be alternating to use this test since <b>any series converges if it converges absolutely. <br /></b>

---+++ Root Test
   $ Statement: <latex>\mbox{ \small Let $\sum u_k$}</latex>%BR%
 be a series with positive terms and%BR%
 <latex>\mbox{ \small $\lim_{k \rightarrow \infty} \sqrt[k]{u_k} = \rho$ }</latex>%BR%
 then:%BR%
 <latex>\left \{ \begin{array} {l} \mbox{ \small Converges if $\rho < 1$ } \\ \mbox{ \small No conclusion if $\rho = 1$ } \\ \mbox{ \small Diverges if $\rho > 1$ } \end{array} \right.</latex>

  Comment: Try this test when <i>u<sub>k</sub></i> involves <i>k<sup>th</sup></i> powers.

---++ Integral Test
   $ Statement: <latex>\mbox{ \small Let $\sum u_k$}</latex>%BR%
 be a series with positive terms and let _f(x)_ be the function that results when
 _k_ is replaced by _x_ in the formula for _u<sub>k</sub>._ %BR%
  If _f_ is a decreasing, continuous function for _x_ > _N_ then:%BR%
 <latex>\mbox{ \small $\sum_{k=N}^{\infty} u_k$ and $\int_{k=N}^{\infty} u_k$ }</latex>%BR%
 have like convergence (either both converge or both diverge).

   $ Comment: Use this test when _f(x)_ is easy to integrate.

---++ Limit Comparison Test
   $ Statement: <latex>\mbox{ \small Let $\sum a_k$ and $\sum b_k$}</latex> be series with positive terms such that:%BR%
    <latex>\mbox{ \small $\lim_{k \rightarrow \infty} \frac{a_k}{b_k} = \rho $}</latex>%BR%
    <latex>\mbox{ \small If $0 < \rho < \infty$}</latex> then the two series have like convergence (either both converge or both diverge).%BR%
    <latex>\mbox{ \small If $\rho = 0$ or $\rho = \infty$}</latex> then notice which series "won".
      * Your unknown series *converges* if it is clearly *smaller than a convergent series* -- think about it.
      * Your unknown series *diverges* if it is clearly *larger than a divergent series* -- think about it.

   $ Comment: This is easier to apply than the _Comparison Test,_ but still requires some skill in choosing the known series. The _Divergence Test_ can be a great source of inspiration here.

---++ Comparison Test
   $ Statement: <latex>\mbox{ \small Let $\sum a_k$ and $\sum b_k$}</latex> be series with positive terms such that:%BR%
    <latex>\mbox{ \small $a_1 \le b_1, a_2 \le b_2, ... a_k \le b_k, ...$}</latex>%BR%
    <latex>\mbox{ \small If $\sum b_k$}</latex> converges then <latex>\mbox{ \small $\sum a_k$}</latex> converges.%BR%
    Similarly, <latex>\mbox{ \small If $\sum a_k$}</latex> diverges, then <latex>\mbox{ \small $\sum b_k$}</latex> diverges.

   $ Comment: <b>Use this test as a last resort.</b> While this test is the foundation of most other tests, other tests are often easier to apply.

---++ Alternating Series Test
   $ Statement: <latex>a_0 - a_1 + a_2 - a_3 + ...</latex>%BR%
 and <latex>- a_0 + a_1 - a_2 + a_3 - ...</latex>%BR%
 or equivalently <latex>\mbox{ \small $\pm \sum_{k=0}^{\infty} (-1)^k a_k$ }</latex>%BR%
 converges, provided <latex>\mbox{ \small $a_0 \ge a_1 \ge a_2 \ge a_3 \ge ... \ge 0$ }</latex>%BR%
 and <latex>\mbox{ \small $\lim_{k \rightarrow \infty} a_k = 0$ }</latex>
   $ Comment: This test applies to *alternating series only.*

---++ Telescoping Series
   $ Statement: Any series where massive cancellation of terms occurs. Often partial sums simplify to a sum of some early terms and some ending terms: everything in between sums to zero (cancels).
   $ Comment: Any time you see individual terms involving funky arithmetic with the indices, be on the lookout for a telescoping series.
      * Break up the typical term into a sum wherever possible.
      * Write out the first few terms.
      * Watch for developing patterns which will allow terms to cancel.

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<latex>\sum_{2}^{\infty} \frac{2}{n^2-1}</latex>

<latex>\mbox{ \small $=\sum_{2}^{\infty} \frac{2}{(n+1)\cdot(n-1)}$}</latex>

<latex>\mbox{ \small $=\sum_{2}^{\infty} \frac{1}{n-1} - \frac{1}{n+1}$ }</latex>

<latex>\mbox{ \small $= \frac{1}{2-1} - \frac{1}{2+1} +\frac{1}{3-1} - \frac{1}{3+1} + \frac{1}{4-1} - \frac{1}{4+1} + \frac{1}{5-1} - \frac{1}{5+1} +\frac{1}{6-1} - \frac{1}{6+1} + ...$}</latex>

<latex>\mbox{ \small $= \frac{1}{1} \textcolor{red}{- \frac{1}{3}} +\frac{1}{2} \textcolor{green}{- \frac{1}{4}} \textcolor{red}{+ \frac{1}{3}} \textcolor{blue}{- \frac{1}{5}} \textcolor{green}{+ \frac{1}{4}} - \frac{1}{6} \textcolor{blue}{+\frac{1}{5}} - \frac{1}{7} + ...$}</latex>

<latex>\mbox{ \small $\sum_{2}^{6} \frac{2}{n^2-1} = \frac{1}{1} +\frac{1}{2} - \frac{1}{6} - \frac{1}{7}$}</latex>

<latex>\mbox{ \small $\sum_{2}^{n} \frac{2}{n^2-1} = \frac{1}{1} +\frac{1}{2} - \frac{1}{n} - \frac{1}{n+1}$}</latex>

<latex>\mbox{ \small $\lim_{n \rightarrow \infty} \sum_{2}^{n} \frac{2}{n^2-1} = 1 +\frac{1}{2} - 0 - 0$}</latex>

<latex>\mbox{ \small $\sum_{2}^{\infty} \frac{2}{n^2-1} = \frac{3}{2}$}</latex>

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-- Main.DickFurnas - 16 Nov 2008
Topic revision: r4 - 2017-10-18 - SteveGaarder