Alternating Series Test

[inkliing]

There's something that's confusing me about what appears to be the standard form of stating the alternating series test in basic calculus. The four sources I looked up were James Stewart's CALCULUS, Howard Anton's CALCULUS, wolfram alpha's mathworld, and wikipedia. All four had essentially the same statement for the alternating series test:

 

If the sum from i=0 to infinity of [(-1)^n][b_n], with b_n>0 for all n, satisfies a) b_(n+1)<=b_n for all n ({b_n} is a decreasing sequence), and B) the limit as n goes to infinity of b_n = 0, then the series is convergent.

 

What I'm confused about is this: since all four sources made it clear that all of the b_n were strictly greater than zero and that b_n->0 as n goes to infinity, what is the point of also adding part a)? Why add that {b_n} must be monotonically decreasing (in the less-than-or-equal-to sense)? It seems to me that b_n>0 for all n and b_n->0 as n goes to infinity implies that {b_n} must be monotonically decreasing (in the less-than-or-equal-to sense).

 

So it seems to me that all four sources should have left out the part about b_(n+1)<=b_n for all n. Am I missing something simple here? It seems to me that the needed counterexample is that of a sequence of real numbers, all greater than zero, which go to zero, but which are not eventually decreasing, which I'm pretty sure is impossible.

 

I would appreciate it if some1 could clearly show that B), coupled with b_n>0 for all n, doesn't imply a) so I can be confident that it is necessary to state a) in the statement of the test.

 

Thanks in advance.

[sanctus]

Your counter-example ;-)

 

sin(x)/x

 

Oscillating not monotone and converging to zero ;-)

[CraigD]

Your counter-example ;-)

 

sin(x)/x

 

Oscillating not monotone and converging to zero ;-)

To this function generate terms of an alternating series, you'd have to define discrete values of x via an iterative formula or a function of the sequence number n, that is:

 

[math] \sum_{n=1}^\infty \frac{\sin\left( f(n) \right) }{f(n)} [/math]

 

A choice of [math]x = f(n) = \left(k +\frac12 \right) \pi[/math] gives

 

[math] \sum_{n=1}^\infty \frac{\sin\left( \left(n +\frac12 \right)\pi \right)}{\left(n +\frac12 \right)\pi} [/math]

 

Which could be simplified to

 

[math]\sum_{n=1}^\infty (-1)^n \frac2{ (2n +1) \pi} [/math]

 

and has a finite value.

 

I can't think of a choice for [math]x = f(k)[/math] that would be the counter-example we're looking for.

 

(de-smilied and relined for readability)

If the sum from i=0 to infinity of [(-1)^n][b_n], with b_n>0 for all n, satisfies

a) b_(n+1)<=b_n for all n ({b_n} is a decreasing sequence), and

B) the limit as n goes to infinity of b_n = 0,

then the series is convergent

...

I would appreciate it if some1 could clearly show that B), coupled with b_n>0 for all n, doesn't imply a) so I can be confident that it is necessary to state a) in the statement of the test.

Here’s a counterexample:

 

[math] \sum_{n=1}^\infty (-1)^{n+1} \left(\frac{2 - (-1)^n}{n} \right) = 1 -1 +\frac13 -\frac12 +\frac15 -\frac13 \dots[/math]

 

Although [math] b_n = \frac{2 - (-1)^n}{n} \to 0 [/math] as [math]n \to \infty[/math], [math]b_4 \not\le b_3[/math], etc.

 

The series fails the alternating series test, so is not assured to, and in fact doesn’t, converge (have a finite value).

Edited April 30, 2011 by CraigD
Fixed typo

[sanctus]

I was thinking that one has to make it depend somehow on an integer, but was writing quickly just assuming that it would work anyway ;-) Was a wrong assumption I see :-)

[sanctus]

Ok, but cheating a bit (i.e not same f(n) in nominator and denominator):

b_n=sin(1/n)/n should do the thing,no? Although it is only defined if one sums from n=1 up...

 

I mean limit is zero and it oscillates.

 

And effectively it does not seem to converge.

[CraigD]

Ok, but cheating a bit (i.e not same f(n) in nominator and denominator):

b_n=sin(1/n)/n should do the thing,no? Although it is only defined if one sums from n=1 up...

…

And effectively it does not seem to converge.

[math]\sum_{n=1}^\infty \frac{\sin \left( \frac1{n}\right)}{n}[/math]

 

is a convergent series, because:

 

[math]sin(a) < a[/math] for all [math]a > 0[/math]

 

[math]\frac{\sin \left( \frac1{n}\right)}{n} < \frac1{n^2}[/math]

 

[math]\sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}{6}[/math] (the famous Basel sequence)

 

It’s not an alternating series, however, because all of its terms have the same, positive, sign, so it fails the alternating series test.

 

Note, importantly, that the alternating series test is a sufficient condition for a series’s convergence, but not a [http://en.wikipedia.org/wiki/Necessary_and_sufficient_condition]necessary[/url] one. Many convergent series fail it, not no series that passes it is divergent.

 

I mean limit is zero and it oscillates.

[math]b_n = \frac{\sin \left( \frac1{n}\right)}{n}[/math] doesn’t oscillate, in the conventional mathematical sense.

 

An oscillating sequence is one that terms don’t converge on a single value, but are always between a finite upper and lower bound (which can be a constant, or a function of n). For example:

 

the sequence [math]b_n = \sin(n)[/math] oscillates between -1 and 1, and doesn’t converge;

 

[math]b_n = \frac{\sin(n)}{n}[/math] oscillates between [math]\frac{-1}{n}[/math] and [math]\frac{1}{n}[/math], and converges on 0.

 

[math]b_n = \frac{\sin \left( \frac1{n}\right)}{n}[/math] doesn’t oscillate, and converges on 0.

[sanctus]

But I used the definition of the OP where he takes the alternating thing outside, i.e let c_n be an alternating series so c_n=(-1)^n*b_n for some b_n. That is why I said my last b_n is an alternating series:-)

[inkliing]

All of your examples are very clear. I understand now why both statements are necessary in the test. Thank you.