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

[Hall]

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?

 

[fresh_42]

You should also use the monotony!

 

[Office_Shredder]

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?

 

[Hall]

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.

 

[Office_Shredder]

You got it. That's the bound you're looking for. Now you just need to finish the proof.

 

[Hall]

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.

 

[Hall]

@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.

 

[Office_Shredder]

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?

 

[Hall]

I think case 2 is a proof in itself, it doesn't use induction hypothesis.

I don't think case 2 works the way you wrote it.

Is my proof wrong?

 

[Office_Shredder]

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.

 

[Hall]

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

for a decreasing positive sequence

 

[Office_Shredder]

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.