Unable to use Cauchy's criterion to prove this Alternating test

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In summary, the given conversation discusses how to show that the expression |s_{N+k} - s_N| is less than anything, for all k, for a given fixed value of N. It is proven that for two terms, the expression is bounded by a_{N+1}, and for three terms, it is also bounded by a_{N+1}. The proof is extended using the induction principle to show that for all terms, the expression is bounded by a_{N+1}. It is then concluded that the sequence of partial sums is Cauchy and therefore converges, thus proving the convergence of the series ##\sum_{n=1}^{\infty} (-1)^{n+1} a
  • #1
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Homework Statement
Given a sequence ##(a_n)## with properties:
1) ##(a_n)## is a decreasing sequence, and ##a_n \gt 0## for all ##n##
2) ##\lim (a_n) = 0##

then the alternating series ##\sum_{n=1}^{\infty} (-1)^{n+1} a_n## converges.
Relevant Equations
To prove the above assertion we're asked to show that the sequence of partial sums
## s_n = a_1 - a_2 + a_3 - a_4 + \cdots (-1)^{n+1} a_n##
is a Cauchy sequence.
For any fixed ##N##, we have
$$
|s_{N+k} - s_N| = | (-1)^{N+2} a_{N+1} + (-1)^{N+3}a_{N+2} + \cdots + (-1)^{N+k+1}a_{N+k} | \lt |a_{N+1}| + |a_{N+2}| + \cdots + |a_{N+k}|
$$
Though, from ##\lim (a_n) = 0## we can establish ## n \geq N \implies |a_n| \lt \varepsilon##, that is all the individual terms in the RHS of the above inequality is less than ##\varepsilon##, yet by increasing ##k## the number of ##\epsilon##'s will increase and hence we won't be able to bound it.

How can we show ##|s_{N+k} - s_N|## to be less than anything, for all k?
 
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  • #2
You should also use the monotony!
 
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  • #3
The alternating is important. You can't throw it out by using the triangle inequality.

Try starting with two terms. Can you find any bound on the size? What about three terms?
 
  • #4
Office_Shredder said:
The alternating is important. You can't throw it out by using the triangle inequality.

Try starting with two terms. Can you find any bound on the size? What about three terms?
With two terms, I can do as follows
$$
| (-1)^{N+2} a_{N+1} + (-1)^{N+3} a_{N+2}| \lt a_{N+1}
$$
Because ##a_{N+1}## is in itself a positive number greater than its successor, and no matter if its prefix be positive or negative, because the next one will be its opposite, thus, diminishing it.

With three terms also,
$$
| (-1)^{N+2} a_{N+1} + (-1)^{N+3}a_{N+2} + (-1)^{N+4} a_{N+3} | \lt a_{N+1}
$$
The reason being that, ##a_{N+2} \gt a_{N+3}##, thus, the prefix sign of ##a_{N+2}## prevails, which is opposite to that of ##a_{N+1}##, hence diminishing it in the end.
 
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  • #5
You got it. That's the bound you're looking for. Now you just need to finish the proof.
 
  • #6
Office_Shredder said:
You got it. That's the bound you're looking for. Now you just need to finish the proof.
Given ##\lim (a_n) = 0##, we have for every ##\epsilon \gt 0## an N such that
$$
n \geq N \implies |a_n| \lt \epsilon$$
as all ##a_n## are positive, we can simply write
$$
n \geq N \implies a_n \lt \epsilon
$$

Consider the alternating series ##\sum_{n=1}^{\infty} (-1)^{n+1} a_n##. Let ##(s_n)## denote sequence of partial sums.
## |s_{N+k} - s_N| = | (-1)^{N+2} a_{N+1} + (-1)^{N+3} a_{N+2} + (-1)^{N+4} a_{N+3} + \cdots + (-1)^{N+k+1} a_{N+k} | \lt a_{N+1} \lt \epsilon##,

(We can prove by induction that for a decreasing positive sequence the following is always true
$$
a_1 - a_2 + a_3 - a_4 \cdots \pm a_n \lt a_1
$$
Base case n=2: ##a_1 -a_2 \lt a_1##. TRUE.

Hypothesis: ##a_1 - a_2 + a_3 - a_4 + \cdots \pm a_n \lt a_1##

Induction:
Case (1): ##a_1 - a_2 + a_3 - a_4 + \cdots + a_n \lt a_1##
It is obvious that, ##a_1 - a_2 + a_3 - a_4 + \cdots + a_n - a_{n+1}\lt a_1##
because ##a_{n+1}## is positive.

Case(2): ##a_1 - a_2 + a_3 - a_4 + \cdots - a_n \lt a_1##
Since, our sequence is decreasing we certainly have
## a_1 + a_3 +a_5+ \cdots+ a_{n+1} \lt a_1 +a_2 + a_4+ \cdots + a_n ##
##a_1 - a_2 +a_3 -a_4 + \cdots -a_n + a_{n+1} \lt a_1##

Hence, proved.)
Hence, the sequence of partial sums is Cauchy, therefore it converges. The series, therefore, ##\sum_{n=1}^{\infty} (-1)^{n+1} a_n ## converges.
 
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  • #7
@Office_Shredder Is my induction correct?

Can we deduce anything about divergence of ##\sum_{n=1}^{\infty} (-1)^{n+1} a_n## from properties of ##a_n##? I don't think it is enough to show that ##(a_n)## increases without bound to prove that the series diverges.
 
  • #8
I don't think case 2 works the way you wrote it. One way to use the inductive hypothesis is to split it out as ##a_1 -(a_2-a_3+...)## and use the inductive hypothesis on the part in parentheses.

Also note just getting ##<a_1## is not enough - how do you know you don't have a large in magnitude negative number?
 
  • #9
I think case 2 is a proof in itself, it doesn't use induction hypothesis.
Office_Shredder said:
I don't think case 2 works the way you wrote it.
Is my proof wrong?
 
  • #10
Oh sorry, case 2 works. My brain couldn't handle ##a_1+a_2+a_4##.

You still have the issue that you have no lower bound on anything.

As far as divergence. Recall the terms must go to zero for a sum to converge. So the only questionable region is if the terms go to zero, but not monotonically. Then it's basically still a mystery, and could go either way.
 
  • #11
Office_Shredder said:
You still have the issue that you have no lower bound on anything
Hall said:
for a decreasing positive sequence
 
  • #12
The only thing you did was write down that the partial sums were ##<a_1##. For a complete proof, you also need to conclude they are bounded from below by some number in order to bound the magnitude.
 

FAQ: Unable to use Cauchy's criterion to prove this Alternating test

What is Cauchy's criterion and how is it used in proofs?

Cauchy's criterion is a mathematical concept used to test the convergence of a series. It states that if the limit of the absolute value of the terms in a series is equal to 0, then the series is convergent. This criterion is often used in proofs to show that a series converges or diverges.

Can Cauchy's criterion be used to prove the convergence of all series?

No, Cauchy's criterion can only be used to prove the convergence of certain types of series, such as absolute convergent series. It cannot be used to prove the convergence of alternating series, which require different methods of proof.

What is the Alternating test and how does it differ from Cauchy's criterion?

The Alternating test is a method used to determine the convergence of an alternating series. It states that if the terms in an alternating series decrease in absolute value and approach 0, then the series is convergent. This test is different from Cauchy's criterion, which only applies to certain types of series.

Why is Cauchy's criterion not applicable to proving the convergence of alternating series?

Cauchy's criterion relies on the absolute value of the terms in a series, which is not useful in proving the convergence of alternating series. Alternating series have terms that alternate between positive and negative, making it impossible to determine the absolute value of the terms.

Are there any other methods that can be used to prove the convergence of alternating series?

Yes, there are other methods that can be used to prove the convergence of alternating series, such as the Leibniz criterion or the Ratio test. These methods take into account the alternating nature of the series and are more suitable for proving their convergence.

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