MathAdmin: /* The Divergence Test */

2016-04-16T16:58:40Z

The Divergence Test

MathAdmin at 18:38, 28 July 2015

2015-07-28T18:38:33Z

MathAdmin: Created page with "'''This page is meant to provide guidelines for actually applying series convergence tests. Although no examples are given here, the requirements for each test are provided.'..."

2015-04-27T16:15:12Z

Created page with "'''This page is meant to provide guidelines for actually applying series convergence tests. Although no examples are given here, the requirements for each test are provided.'..."

New page

'''This page is meant to provide guidelines for actually applying series convergence tests. Although no examples are given here, the requirements for each test are provided.'''

== Important Series ==

There are two series that are important to know for a variety of reasons. In particular, they are useful for comparison tests. 

'''Geometric series.''' These are series with a common ratio <math style="vertical-align: 0%">r</math> between adjacent terms which are usually written 

::<math>\sum_{k=0}^{\infty}a_{0}r^{k}.</math>

These are convergent if <math style="vertical-align: -22%">|r|<1</math>, and divergent if <math style="vertical-align: -22%">|r|\geq1</math>. If it is convergent, we can find the sum by the formula 

::<math>S=\frac{a_{0}}{1-r},</math> 

where <math style="vertical-align: -12%">a_{0}</math> is the first term in the series (if the index starts at <math style="vertical-align: 0%">k=2</math> or <math style="vertical-align: 0%">k=6</math>, then "<math style="vertical-align: -12%">a_{0}</math>" is actually the first term <math style="vertical-align: -12%">a_{2}</math> or <math style="vertical-align: -12%">a_{6}</math>, respectively).

'''''p''-series.''' These are series of the form

::<math>\sum_{k=1}^{\infty}\frac{1}{k^{p}}.</math>

If <math style="vertical-align: -15%">p>1</math>, then the series is convergent. On the other hand, if <math style="vertical-align: -15%">p\leq1</math>, the ''p''-series is divergent.

== The Divergence Test ==

If <math style="vertical-align: -65%">{\displaystyle \lim_{k\rightarrow\infty}a_{k}\neq0,}</math> then the series/sum <math style="vertical-align: -98%">\sum_{k=0}^{\infty}a_{k}</math> diverges.

'''Note:''' The opposite result ''doesn't'' allow you to conclude a series converges. If <math style="vertical-align: -60%">{\displaystyle \lim_{k\rightarrow\infty}a_{k}=0}</math> , it merely indicates the series ''might'' converge, and you still need to confirm it through another test.

In particular, the sequence <math style="vertical-align: -38%">\left\{ \frac{1}{k}\right\} </math> converges to zero, but the sum <math style="vertical-align: -65%">\sum_{k=0}^{\infty}\frac{1}{k}</math> , our harmonic series, diverges.

== The Integral Test ==

Suppose the function <math style="vertical-align: -20%">f(x)</math> is continuous, positive and decreasing on some interval <math style="vertical-align: -22%">[c,\infty)</math> with <math style="vertical-align: -13%">c\geq1</math>,
and let <math style="vertical-align: -21%">a_{k}=f(k)</math>. Then the series <math style="vertical-align: -87%">\sum_{k=b}^{\infty}a_{k}</math> is convergent if and only if <math style="vertical-align: -13%">c\geq b</math> and

::<math>\int_{c}^{\infty}f(x)\, dx</math>

is convergent (not infinite).

'''Note:''' This test, like many of them, has a few specific requirements. In order to use it on a test, you need to state/show:

:* For all <math style="vertical-align: -15%">k\geq c</math> for some <math style="vertical-align: -15%">c\geq b</math>, the function is positive. (Most of the time, <math style="vertical-align: 0%">c</math> is just my starting index <math style="vertical-align: 0%">b</math>).
:* For all <math style="vertical-align: -15%">k\geq c</math>, the function is decreasing.
:* The integral is convergent (or divergent, if you're proving divergence).

''Then,'' you can say, "By the Integral Test, the series is convergent (or divergent)."

I wrote this with <math style="vertical-align: 0%">c</math> instead of <math style="vertical-align: 0%">b</math> for a lower bound to indicate ''you only need to show the series and function are "eventually" decreasing, positive, etc''. In other words, we don't care what happens at the beginning (or head) of a series - only at the end (or tail).

== The Comparison Test ==

Suppose <math style="vertical-align: -98%">\sum_{k=1}^{\infty} b_{k}</math> is a series ''with positive terms'', and <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> is a series ''with eventually positive terms''. Then

:* If for all <math style="vertical-align: -15%">k\geq c</math> for some <math style="vertical-align: 0%">c</math> greater than or equal to our starting index, and <math style="vertical-align: -98%">\sum_{k=1}^{\infty} b_{k}</math> is convergent, then <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> is convergent.

:* If <math style="vertical-align: -15%">a_{k}\geq b_{k}</math> for all <math style="vertical-align: 0%">k</math> and <math style="vertical-align: -98%">\sum_{k=1}^{\infty} b_{k}</math> is divergent, then <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> is divergent.

'''Note:''' Requirements for this test include showing (or at least stating):

:* For all <math style="vertical-align: -15%">k\geq c</math> for some <math style="vertical-align: 0%">c</math> greater than or equal to our starting index, <math style="vertical-align: -15%">a_{k}</math> is positive. (Most of the time, <math style="vertical-align: 0%">c</math> is just the starting index.)

:* For all <math style="vertical-align: -15%">k\geq c</math>, <math style="vertical-align: -15%">a_{k}\leq b_{k}</math> for convergence, or <math style="vertical-align: -15%">a_{k}\geq b_{k}</math> for divergence.
:* ''(This is important)'' State why <math style="vertical-align: -98%">\sum_{k=1}^{\infty} b_k</math> is convergent, such as a ''p''-series with <math style="vertical-align: -20%">p>1</math>, or a geometric series with <math style="vertical-align: -24%">|r|<1</math>. Obviously, you would need to state why it is divergent if you're showing it's divergent.

''Then'', you can say, "By the Comparison Test, the series is convergent (or divergent)."

== The Limit Comparison Test ==

Suppose <math style="vertical-align: -100%">\sum_{k=1}^{\infty} a_{k}</math> and <math style="vertical-align: -100%">\sum_{k=1}^{\infty} b_{k}</math>
are series with positive terms. If <math style="vertical-align: -75%">\lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}=c</math> where <math style="vertical-align: -5%">0<c<\infty</math>, then either both series converge, or both series diverge.

Additionally, if <math style="vertical-align: 0%">c=0</math> and <math style="vertical-align: -100%">\sum_{k=1}^{\infty} b_{k}</math> converges, <math style="vertical-align: -100%">\sum_{k=1}^{\infty} a_{k}</math> also converges. Similarly, if <math style="vertical-align: -5%">c=\infty</math>  and <math style="vertical-align: -100%">\sum_{k=1}^{\infty} b_{k}</math> diverges, then <math style="vertical-align: -100%">\sum_{k=1}^{\infty} a_{k}</math> also diverges.

'''Note''': First of all, let's mention the fundamental idea here. If some series <math style="vertical-align: -100%">\sum_{k=1}^{\infty} b_{k}</math> converges, then <math style="vertical-align: -100%">\sum_{k=1}^{\infty} cb_{k}</math> converges where <math style="vertical-align: -22%">c\neq\pm\infty </math> is a constant. This test shows that one series ''eventually'' is just like the other one multiplied by a constant, and for that reason it will also converge/diverge
if the one compared to converges/diverges. To use it, you need to state/show:

:*<math style="vertical-align: -15%">a_{k} </math> is eventually positive (really, non-negative).

:*<math style="vertical-align: -72%">{\displaystyle \lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}}=c </math>.

:* State why <math style="vertical-align: -100%">\sum_{k=1}^{\infty} b_{k} </math> is convergent, such as a ''p''-series with <math style="vertical-align: -20%">p>1 </math>, or a geometric series with <math style="vertical-align: -20%">|r|<1 </math>. Obviously, you would need to state why it is divergent if you're showing it's divergent.
''Then'', you can say, "By the Limit Comparison Test, the series is convergent (or divergent)."

Like the Comparison Test and the Integral Test, it's fine if the first terms are kind of "wrong" - negative, for example - as long as they eventually wind up (for <math style="vertical-align: 0%">k>c</math> for a particular <math style="vertical-align: 0%">c</math> ) meeting
the requirements.

== The Alternating Series Test ==

If a series <math>\sum_{k=1}^{\infty} a_{k}</math> is

:*Alternating in sign, and

:*<math>\lim_{k\rightarrow 0}|a_{k}|=0,</math>

then the series is convergent.

'''Note:''' This is a fairly straightfoward test. You only need to do two things:

:*Mention the series is alternating (even though it's usually obvious).

:*Show the limit converges to zero.

''Then'', you can say, "By the Alternating Series Test, the series is convergent."

As an additional detail, if it fails to converge to zero, then you would say it diverges by the Divergence Test, ''not'' the Alternating Series Test.

== The Ratio Test ==

Let <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> be a series. Then:

:*If <math style="vertical-align: -84%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L<1</math>, the series is absolutely convergent (and therefore convergent).

:*If <math style="vertical-align: -84%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L>1</math> or <math style="vertical-align: -83%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=\infty</math>, the series is divergent.

:*If <math style="vertical-align: -84%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=1</math>, the Ratio Test is inconclusive.

'''Note:''' Both this and the Root Test have the least requirements. The Ratio Test ''does'' require that such a limit exists, so a series like

<math>0+1+0+\frac{1}{4}+0+\frac{1}{9}+\cdots</math>

could not be assessed as written with the Ratio Test, as division by zero is undefined. You might have to argue it's the same sum as

<math>1+\frac{1}{4}+\frac{1}{9}+\cdots,</math>

and you could then apply the Ratio Test.

== The Root Test ==

Let <math style="vertical-align: -91%">\displaystyle\sum_{k=0}^{\infty} a_{k}</math> be a series. Then:

:*If <math style="vertical-align: -64%">{\displaystyle \lim_{k\rightarrow\infty}\sqrt[k]{|a_{k}|}}=L<1,</math> the series is absolutely convergent (and therefore convergent).

:*If <math style="vertical-align: -64%">{\displaystyle \lim_{k\rightarrow\infty}\sqrt[k]{|a_{k}|}}=L>1</math> or <math style="vertical-align: -64%">{\displaystyle\lim_{k\rightarrow\infty}\sqrt[k]{|a_{k}|}}=L=\infty,</math>
the series is divergent.
:*If <math style="vertical-align: -64%">{\displaystyle \lim_{k\rightarrow\infty}\sqrt[k]{|a_{k}|}}=L=1</math>, the Root Test is inconclusive.

← Older revision		Revision as of 16:58, 16 April 2016
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	== The Divergence Test ==		== The Divergence Test ==

−	If <math style="vertical-align: -65%">{\displaystyle \lim_{k\rightarrow\infty}a_{k}\neq0,}</math> then the series/sum <math style="vertical-align: -98%">\sum_{k=0}^{\infty}a_{k}</math> diverges.	+	If <math style="vertical-align: -65%">{\displaystyle \lim_{k\rightarrow\infty}a_{k}\neq0,}</math> then the series/sum <math style="vertical-align: -98%">\sum_{k=0}^{\infty}a_{k}</math>  diverges.

← Older revision		Revision as of 18:38, 28 July 2015
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	the series is divergent.		the series is divergent.
	:*If <math style="vertical-align: -64%">{\displaystyle \lim_{k\rightarrow\infty}\sqrt[k]{\|a_{k}\|}}=L=1</math>, the Root Test is inconclusive.		:*If <math style="vertical-align: -64%">{\displaystyle \lim_{k\rightarrow\infty}\sqrt[k]{\|a_{k}\|}}=L=1</math>, the Root Test is inconclusive.
		+
		+
		+	'''Contributions to this page were made by [[Contributors\|John Simanyi]]'''

Series - Tests for Convergence/Divergence - Revision history

MathAdmin: /* The Divergence Test */

MathAdmin at 18:38, 28 July 2015

MathAdmin: Created page with "'''This page is meant to provide guidelines for actually applying series convergence tests. Although no examples are given here, the requirements for each test are provided.'..."