Convergence of Series with Decreasing Terms: Can na_n Tend to Zero?

[Treadstone 71]

"Suppose $$a_1\geq a_2\geq ... \geq a_n > 0$$ for all $$n$$ such that $$\sum a_n$$ converges. Show that $$na_n$$ tends to zero."

The best I can do is prove that a subsequence $$na_n$$ converges to a number between the sum of the series and 0.

 

[NateTG]

"Suppose $$a_1\geq a_2\geq ... \geq a_n > 0$$ for all $$n$$ such that $$\sum a_n$$ converges. Show that $$na_n$$ tends to zero."

The best I can do is prove that $$na_n$$ converges to a number between the sum of the series and 0.

Have you tried assuming that $na_n$ does not tend to zero and using a proof by contradiction?

 

[AKG]

Since the series converges, a_n converges to zero. If na_n doesn't converge to 0, then that means that a_n converges to zero no faster than 1/n. But if a_n converges no faster than 1/n, then the series will be no "better" than than the harmonic series, and so it certainly will not converge - contradiction. This is the rough idea you might want to work with, but of course, it's not rigorous yet.

 

[Treadstone 71]

I thought about that approach, but the problem is that if it converges to a number less than 1, then I can't say that it the term a_n converges no faster than the harmonic series.

 

[AKG]

I thought about that approach, but the problem is that if it converges to a number less than 1, then I can't say that it the term a_n converges no faster than the harmonic series.

You can use the same idea. If it converges to some number c, such that 1 > c > 0, then a_n converges no faster than c/n, which means the series is no better than:

$$\sum \frac{c}{n} = c\sum \frac{1}{n}$$

 

[Treadstone 71]

I've gotten so far:

na_n is bounded, therefore has a convergent subsequence.

IF na_n converges, then it converges to 0.

All I have to do now is prove that na_n converges, which I am unable to.

 

[NateTG]

Let's say you have some $n_1$
such that
$$n_1a_{n_1} > \epsilon$$

Can you give a (non-zero) lower bound on
$$\sum_{i=\frac{n_1}{2}}^{n_1} a_i$$

Now, if, in addition, you had
$$n_2 > 2 \times n_1$$
with
$$n_2a_{n_2} > \epsilon$$
can you give a lower bound on
$$\sum_{i=\frac{n_1}{2}}^{n_2} a_i$$

 

[Treadstone 71]

The lower bound for the first one would be the difference $$n_1a_n_1 - n_1/2 a_{n_1/2}$$ assuming n1 is even. I don't see how this helps. Are you trying to show that na_n is Cauchy?

 

[quasar987]

For an easy plug & chug way: apply the D'Alembert criterion (aka ratio test) on na_n.

 

[Treadstone 71]

The ratio test is for the series na_n, not the sequence na_n. The fact is that there exists converging a_n such that $$\sum a_n$$ converges but $$\sum na_n$$ does not. 1/n^2, for instance.

 

[AKG]

CASE 1:
Suppose {na_n}_{n in N} converges to some finite c > 0. Suppose that for all N > 0, there exists n > N such that na_n < c/2. Well then {na_n}_{n in N} could not converge to c, countradicting our assumption. So there is some N > 0 such that for all n > N, na_n > c/2. Thus we can choose a subsequence {na_n}_{n > N} which has each entry is at least c/2.

But then

$$\sum _{n \in \mathbb{N}} a_n = \sum _{n \in \mathbb{N}} \frac{na_n}{n} \geq \sum _{n > N} \frac{na_n}{n} \geq \sum _{n > N} \frac{c}{2n} = \frac{c}{2} \sum _{n > N}\frac{1}{n}$$

But if the sum on the left converges, then it cannot be greater than the sum on the right, because the one on the right clearly diverges. Contradiction.

CASE 2:
{na_n}_{n in N} converges to some negative number. Impossible, since all the terms are positive. Contradiction.

CASE 3:
{na_n}_{n in N} diverges. Same argument as in Case 1. Contradiction.

CASE 4:
{na_n}_{n in N} converges to 0. By process of elimination, the only posibility, thus it must be the case. Q.E.D.

 

[NateTG]

The lower bound for the first one would be the difference $$n_1a_n_1 - n_1/2 a_{n_1/2}$$ assuming n1 is even. I don't see how this helps. Are you trying to show that na_n is Cauchy?

Let me try it this way:
What is the minimum value of a term in the sum.
How many terms are there?

 

[Treadstone 71]

CASE 1:
Suppose {na_n}_{n in N} converges to some finite c > 0. Suppose that for all N > 0, there exists n > N such that na_n < c/2. Well then {na_n}_{n in N} could not converge to c, countradicting our assumption. So there is some N > 0 such that for all n > N, na_n > c/2.

If it converges to any number, then I can prove that number must be 0:

$$na_n \rightarrow c > 0 \Rightarrow \frac{na_n}{n\frac{1}{n} c}=\frac{a_n}{\frac{c}{n}} \rightarrow 1 \Rightarrow \sum a_n, \sum\frac{c}{n}$$ converge or diverge simultaneously. Since $$\sum\frac{c}{n}$$ diverges, so does $$\sum a_n$$, which is impossible.

However, $$na_n$$ need not converge! All we know is that it's bounded, and therefore by Bolzano-Weierstrass theorem, there exists a converging subsequence. However, the above proof, using the converging subsequence, fails, since $$\sum\frac{c}{n_k}$$ might be convergent, and thus not leading into a contradiction.

The problem with your proof is that I can't see how you can conclude for all n>N, since it should be a subsequence only.

 

[NateTG]

How's this:

Assume, by contradiction that
$$\lim_{n \rightarrow \infty} na_n \neq 0$$

Then there is some $\epsilon > 0$ such that for any $n \in \mathbb{N}$ there is some $N > n$ such that $$Na_N > \epsilon$$.

So, clearly there is an $n_1>3$ so that $n_1a_{n_1}>\epsilon$, and for any $i>1$ there is some $n_i$ such that $n_i>2\times n_i-1$ and $n_ia_{n_i}>\epsilon$.

Now $\sum_{j=\ciel{\frac{n_i}{2}}}^{n_i} a_n > \frac{\epsilon}{4}$
since the sum contains at least $\frac{n_i}{4}$ terms all greater than $\frac{\epsilon}{n_i}$.

Since $n_i<\frac{n_{i+1}}^{2}$ we have $n_i<\ceil(\frac{n_{i+1}}{2})$

Now, we can regoup
$$\sum_{j=1}^{\infty}a_j \geq \sum_{i=0}^{\infty} \sum_{j=\ciel{\frac{n_i}{2}}}^{n_i} a_j > \sum_{i=0}^{\infty} \frac{\epsilon}{4}$$

But this contradicts the assumptiont that the sum is convergent.

Hence the limit must be equal to zero.

 

[AKG]

However, $$na_n$$ need not converge!

Which is why I gave case 3.

All we know is that it's bounded

What do you mean "all we know is that it's bounded." A priori, it might be unbounded, as it could be that a_n = 1. On the other hand, since we're asked to prove that {na_n} converges to 0, we "know" that not only is it bounded, but that it converges to 0.

The problem with your proof is that I can't see how you can conclude for all n>N, since it should be a subsequence only.

No, I started with:

Suppose that for all N > 0, there exists n > N such that na_n < c/2

which is:

$$(\forall N > 0)(\exists n > N)(na_n < c/2)$$

I derived a contradiction from this, so I may infer its negation, which is:

$$(\exists N > 0)(\forall n > N)(na_n \geq c/2)$$

which is:

there is some N > 0 such that for all n > N, na_n > c/2.

 

[Treadstone 71]

I know it's bounded because $$\sum_{1}^{\infty}a_i \geq \sum_{1}^{n}a_i \geq na_n > 0$$ since a_n is decreasing.

 

[NateTG]

CASE 1:
Suppose {na_n}_{n in N} converges to some finite c > 0. Suppose that for all N > 0, there exists n > N such that na_n < c/2. Well then {na_n}_{n in N} could not converge to c, countradicting our assumption. So there is some N > 0 such that for all n > N, na_n > c/2. Thus we can choose a subsequence {na_n}_{n > N} which has each entry is at least c/2.
...
CASE 3:
{na_n}_{n in N} diverges. Same argument as in Case 1. Contradiction.

I don't see how the argument from case 1 applies to case 3 since there is, by definition, no well-defined limit for a divergent sequence. Moreover, there are divergent sequences where there is no suitable $c$.

For example, consider that if $a_n=\frac{||-1^n+1|+.5^n}{n}$, then $\left{na_n\right}$ is divergent since it has subsequences that converge to 0 and 2 (the odd and even terms respectively). For any finite $c>0$ and any $n \in \mathbb{N}$, there is some $n>N$ so that $na_n<\frac{c}{2}$ since the odd terms converge to zero.

I

 

[AKG]

Sorry, case 1 doesn't exactly apply to case 3. It does to the subcase that the sequence diverges to infinity. In this case, pick any finite c > 0, and the same argument applies, but you'd replace "could not converge to c" in the second sentence with "could not diverge to infinity". However, case 1 doesn't apply to the subcase where the sequence neither converges nor diverges to infinity, but is simply bounded divergent (i.e., it "oscillates"). Note that Treadstone proved that the sequence is bounded, so we don't really need to apply case 1 to the first subcase of case 3 (the subcase where the sequence diverges to infinity), we can just say that this subcase will never happen

So, what to do when the sequence oscillates? Well, observe that:

$$\lim _{n \to \infty}\frac{(n+1)a_{n+1}}{na_n} = \lim _{n \to \infty}\frac{n+1}{n}\lim _{n \to \infty}\frac{a_{n+1}}{a_n} = \lim _{n \to \infty}\frac{a_{n+1}}{a_n} \leq 1$$

since {a_n} converges, which is true since the series converges. At least, I think that's correct. If it is, then I think it can be used to argue that {na_n} in fact does not oscilate, thereby eliminating the remaining subcase of case 3.

 

[NateTG]

A cleaner approach

$$na_n = \sum_{i=1}^{n} {a_n} \leq 2 \sum_{i=\lfloor \frac{n}{2} \rfloor}^{n}{a_n} \leq \sum_{i=\lfloor \frac{n}{2} \rfloor}^{\infty}{a_i} \leq \sum_{i=\lfloor\frac{n}{2} \rfloor}^{\infty}{a_i}$$

The $\leq$ are justified because $a_n$ is strictly positive, $a_n$ is monotone decreasing, and $a_n$ is strictly positive respectively.
Now we have
$$0 < na_n \leq \sum_{i=\lfloor\frac{n}{2} \rfloor}^{\infty}{a_i}$$

Since the sequence converges, the limit on the R.H.S. as n goes to infinity must be zero. This bounds $na_n$ above and below with sequences that go to zero, therefore $na_n$ goes to zero.

 

[AKG]

I think you mean:

$$na_n\ =\ \sum_{i=1}^{n} {a_n}\ \leq\ 2\left ( \sum_{i=\lfloor \frac{n}{2} \rfloor}^{n}{a_n}\right )\ \leq\ 2\left ( \sum_{i=\lfloor \frac{n}{2} \rfloor}^{n}{a_i}\right )\ \leq\ 2\left ( \sum_{i=\lfloor\frac{n}{2} \rfloor}^{\infty}{a_i}\right )$$

Then applying the squeeze theorem (i.e. your argument) to:

$$0\ <\ na_n\ \leq\ 2\left ( \sum_{i=\lfloor\frac{n}{2} \rfloor}^{\infty}{a_i}\right )$$

we get the desired result. Very nice!

 

[Treadstone 71]

Very clever. Thanks for the help!