Sequence/Series proof

[STEMucator]

Homework Statement

Let $\sum a_n$ be a convergent series of non-negative terms and suppose that the sequence $\{a_n\}$ is non-increasing.

Apply the Cauchy Criterion to show that $\forall ε > 0$, if m and n are sufficiently large, then $na_n < (ε/2) + ma_n$.

Hint : $a_{m+1} + ... + a_n ≥ (n-m)a_n$ for $n>m$.

Use this fact ( fix m ) to show that $na_n → 0$ as $n → ∞$.

Give an example of a series where $na_n → 0$ as $n → ∞$, $\{a_n\}$ is not increasing and yet $\sum a_n$ diverges.

Homework Equations

Cauchy Criterion :

A sequence $\{a_n\}$ converges $⇔ \forall ε > 0, \exists N \space | \space n,m > N \Rightarrow |a_n - a_m| < ε$.

The Attempt at a Solution

First time I've seen something like this. I'm not sure what the question is asking me to do in particular. Then again I'm not quite sure where to start this one?

So if $na_n < (ε/2) + ma_n$, then $(n-m)a_n < ε/2$. I have a feeling the hint comes into play somehow here, but I'm not seeing how it all comes together here. ( I'm thinking I should assume W.L.O.G, that n>m ).

Also as for the example portion of the question, $\{a_n\} = \{ \frac{1}{n^{3/2}} \}$ works just fine I can see that.

If anyone could help me out a bit it would be much appreciated.

 

[pasmith]

You need to apply the Cauchy criterion to the sequence of partial sums $S_n = \sum_{k=0}^n a_k$. Without loss of generality you can assume that n > m.

(The only twist is that instead of requiring N such that $|S_n - S_m| < \epsilon$, you must instead require N such that $|S_n - S_m| < \epsilon/2$.)

EDIT: by the integral test, $\sum n^{-3/2}$ converges.

 

[STEMucator]

You need to apply the Cauchy criterion to the sequence of partial sums $S_n = \sum_{k=0}^n a_k$. Without loss of generality you can assume that n > m.

(The only twist is that instead of requiring N such that $|S_n - S_m| < \epsilon$, you must instead require N such that $|S_n - S_m| < \epsilon/2$.)

EDIT: by the integral test, $\sum n^{-3/2}$ converges.

Ohhh sigma a_n diverges, I missed that part. Okay I'll think of another example afterwards.

As for the main question, are you trying to say I should assume W.L.O.G that n>m and let :

$s_n = \sum_{k=0}^{n} a_k$ and $s_m = \sum_{j=0}^{m} a_j$

So that by the Cauchy Criterion, choosing N = ε/2 :

$|s_n - s_m| = \sum_{k=0}^{n} a_k - \sum_{j=0}^{m} a_j < ε/2$

Is this on the right track?

 

[pasmith]

Ohhh sigma a_n diverges, I missed that part. Okay I'll think of another example afterwards.

As for the main question, are you trying to say I should assume W.L.O.G that n>m and let :

$s_n = \sum_{k=0}^{n} a_k$ and $s_m = \sum_{j=0}^{m} a_j$

So that by the Cauchy Criterion, choosing N = ε/2 :

$|s_n - s_m| = \sum_{k=0}^{n} a_k - \sum_{j=0}^{m} a_j < ε/2$

Is this on the right track?

Yes. But you are choosing N such that if $n > m > N$ then $|s_n - s_m| < \epsilon/2$.

Now use the fact that n > m (so that $s_n$ includes all the terms of $s_m$ plus some extra terms) to work out the difference between the two.

 

[STEMucator]

Yes. But you are choosing N such that if $n > m > N$ then $|s_n - s_m| < \epsilon/2$.

Now use the fact that n > m (so that $s_n$ includes all the terms of $s_m$ plus some extra terms) to work out the difference between the two.

Okay I'll attempt to formalize a bit in this post and collect thoughts and actually try to get something going.

So suppose without loss of generality for the duration of this question that n > m, i.e n - m > 0.

So by the C.C, we want to show $\forall ε>0, \exists N \space | \space m > n > N \Rightarrow |s_n - s_m| < ε$

Let : $s_n = \sum_{k=1}^{n} a_k$ and $s_m = \sum_{k=1}^{m} a_k$

Since m > n, we know that s_n contains all the terms of s_m plus extra left over terms. To find these other terms we consider the following :

Let j = n - m so that j + m = n, then we have :

$|s_{j+m} - s_m| ≤ |s_j| ≤ |a_{1+m} + ... + a_{j+m}|$

^^ I'm getting a bit stuck here presuming I'm on the right track. I'm not sure how to continue this train of thought.

 

[STEMucator]

Wait a second... a thought occurred to me a moment ago. I'm not really sure it has any merit (since pas got off), but I will post it and see what anyone who views this may think.

Suppose without loss of generality for the duration of this question that n > m.

Let the sequences of partial sums be defined as : $s_n = \sum_{k=1}^{n} a_k$ and $s_m = \sum_{k=1}^{m} a_k$

We know that $\sum a_n$ converges which implies $a_n → 0$ as $n → ∞$. Hence the sequences of partial sums also converge.

So by the C.C, we want to show $\forall ε>0, \exists N \space | \space n > m > N \Rightarrow |s_n - s_m| < ε$

Since m > n, we know that $s_n$ contains all the terms of $s_m$ plus extra left over term(s). To find these other terms we consider the following :

Let k = n - m so that k + m = n, then we have :

$|s_{m+k} - s_m| ≤ |s_k| = |a_{m+1} + ... + a_{n}| ≤ a_{m+1} + ... + a_{n}$

Now from the hint, we know that $a_{m+1} + ... + a_{n} ≥ (n-m)a_n$.

If $a_{m+1} + ... + a_{n} = (n-m)a_n$, then trivially we have $|s_{m+k} - s_m| ≤ |s_k| < ε/2$ as desired. Otherwise if $a_{m+1} + ... + a_{n} > (n-m)a_n$, then we have a contradiction because we stated earlier that n > m.

Also, for showing $na_n → 0$, fix m=0 as suggested. Since $\sum a_n$ converges, once again as stated before, $a_n → 0$ as $n → ∞$. So multiplying $a_n$ by an integer will not affect the result. Thus $na_n → 0$ as $n → ∞$.

As for the example take : $a_n = \frac{(-1)^n}{n}$.

 

[pasmith]

Wait a second... a thought occurred to me a moment ago. I'm not really sure it has any merit (since pas got off), but I will post it and see what anyone who views this may think.

Suppose without loss of generality for the duration of this question that n > m.

Let the sequences of partial sums be defined as : $s_n = \sum_{k=1}^{n} a_k$ and $s_m = \sum_{k=1}^{m} a_k$

We know that $\sum a_n$ converges which implies $a_n → 0$ as $n → ∞$.

This is true.

the sequences of partial sums also converge.

This is the definition of convergence of an infinite series.

So by the C.C, we want to show $\forall ε>0, \exists N \space | \space n > m > N \Rightarrow |s_n - s_m| < ε$

No. We already know this: we are given that $\sum a_n$ converges. Thus by definition the sequence of partial sums converges, and the Cauchy criterion then says that for all $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that if $n > m \geq N$ then $|s_n - s_m| < \frac12\epsilon$.

Since m > n, we know that $s_n$ contains all the terms of $s_m$ plus extra left over term(s). To find these other terms we consider the following :

Let k = n - m so that k + m = n, then we have :

$|s_{m+k} - s_m| ≤ |s_k| = |a_{m+1} + ... + a_{n}| ≤ a_{m+1} + ... + a_{n}$

The first relation is true (at least under the assumption that the $a_n$ are non-negative and non-increasing) but unhelpful. The next relation is false: $s_k = a_1 + \cdots + a_k$. The final relation is false unless all the terms are non-negative.

It ought to be obvious that if $m < n$ then
$$\sum_{k=1}^n a_k = \sum_{k=1}^m a_k + \sum_{k=m+1}^n a_k$$
so that
$$|s_n - s_m| = \left|\sum_{k=1}^n a_k - \sum_{k=1}^m a_k\right| = \left|\sum_{k=m+1}^n a_k\right| = |a_{m+1} + \cdots + a_{n}| = a_{m+1} + \cdots + a_{n}$$
with the last equality following because the terms are non-negative.

Now from the hint, we know that $a_{m+1} + ... + a_{n} ≥ (n-m)a_n$.

So, putting the hint together with the above, we have that for all $\epsilon > 0$ and $n > m$ sufficiently large,
$$(n - m)a_n \leq a_{m+1} + \dots + a_n = |s_n - s_m| < \textstyle \frac12 \epsilon$$
from which it follows that
$$na_n < \textstyle\frac12 \epsilon + ma_n$$
which was what the first part of the question asked you to prove.

Also, for showing $na_n → 0$, fix m=0 as suggested.

You can't fix $m = 0$. The assumption we're operating under is that $n > m \geq N$ where $N$ is given by the Cauchy criterion as above.

Since $\sum a_n$ converges, once again as stated before, $a_n → 0$ as $n → ∞$. So multiplying $a_n$ by an integer will not affect the result. Thus $na_n → 0$ as $n → ∞$.

I'm not sure what you're trying to say here. If you're saying that for fixed $m$, $ma_n\to 0$ since $a_n \to 0$ then you are right, but that only gets you
$$0 \leq \lim_{n \to \infty} na_n \leq \textstyle\frac12 \epsilon$$
and you need to explain why that requires that $na_n \to 0$. If you're saying that $na_n \to 0$ if $a_n \to 0$ (which if you're assuming m = 0 might be the case) then that is simply false.

As for the example take : $a_n = \frac{(-1)^n}{n}$.

This doesn't satisfy the condition that $na_n \to 0$.

Finding examples is almost impossible if you don't know where to look, so I'll just suggest that you look at $(n\ln n)^{-1}$ for $n \geq 2$ (it really doesn't matter what you define $a_1$ to be as long as $a_1 \geq (2\ln 2)^{-1}$ so the sequence is non-increasing).

The point of the question is that you've proven that if $\sum a_n$ converges and the terms are non-negative and non-increasing, then $na_n \to 0$. However the last part of the question shows that the converse does not hold: the fact that $na_n \to 0$ does not imply that $\sum a_n$ converges.