May | 2023 | Tao Analysis Solutions

Exercise statement

Prove Proposition 7.2.5.

Proposition 7.2.5. Let $\sum_{n=m}^\infty a_n$ be a formal series of real numbers. Then $\sum_{n=m}^\infty a_n$ converges if and only if, for every real number $\varepsilon > 0$ , there exists an integer $N \geq m$ such that

$\displaystyle \left|\sum_{n=p}^q a_n\right| \leq \varepsilon$ for all $p,q \geq N$ .

Hints

Use Proposition 6.1.12 and Theorem 6.4.18.
If you want to do this exercise correctly, it is not as easy as just applying the given Proposition and/or Theorem!

How to think about the exercise

If you look up the hints, it is quite obvious how the exercise is done. I wanted to give a thought process of how to approach the problem if one randomly saw it (i.e. without accompanying hints), so in what follows I won’t be assuming that the hints have been given.

Let’s write out the two sides of the “if and only if” by making the quantifiers explicit:

For all $\varepsilon > 0$ there exists $N \geq m$ such that for all $p,q \geq N$ we have $\left|\sum_{n=p}^q a_n\right| \leq \varepsilon$ (1)

The claim that $\sum_{n=m}^\infty a_n$ converges is:

For all $\varepsilon > 0$ there exists $N \geq m$ such that for all $k \geq N$ we have $\left|\sum_{n=m}^k a_n - S\right| \leq \varepsilon$ (2)

Above, I have used $S$ to denote the sum. It’s actually equal to $S=\sum_{n=m}^\infty a_n$ .

This shows something interesting. When formalizing what it meant for a series to converge, i.e., when writing out (2), we had to refer to what the sum converges to, i.e., to $S$ . But in (1), we did not even have to refer to the final sum! In other words, (1) encodes the claim “this series converges” without talking about what the series converges to. This sounds awfully familiar … didn’t we have a way to say “this sequence converges” without referring to the limit of the sequence? That’s right, this is sounding exactly like what Cauchy sequences are. And in fact, if our pattern-matching is active, we might even notice that in (1) we had “for all $p,q \geq N$ ” which is reminiscent of the “for all $j,k \geq N$ ” that we used a lot when dealing with Cauchy sequences (the variable names don’t matter of course).

So let’s see … if $S_N = \sum_{n=m}^N a_n$ is the $N$ th partial sum of the series, then $\sum_{n=p}^q a_n$ is basically equal to $S_q - S_p$ . (How did I see this? It was so quick it’s hard for me to say what my mind did here. I suppose it’s pattern-matching on things I’ve seen before, e.g. the integral rule $\int_a^b + \int_b^c = \int_a^c$ , which can be written in subtracted form as $\int_a^c - \int_a^b =\int_b^c$ .) So $\left|\sum_{n=p}^q a_n\right| \leq \varepsilon$ is basically saying $|S_q - S_p| \leq \varepsilon$ . So (1) is basically saying that $(S_N)_{N=m}^\infty$ is Cauchy. And we have exactly the theorems to show that “is Cauchy” and “is convergent” are the same for sequences.

I said “basically” above because $\sum_{n=p}^q a_n = S_q - S_p$ only holds when $q \geq p$ . When $q < p$ , the sum $\sum_{n=p}^q a_n = 0$ whereas $S_q - S_p$ is not necessarily zero; instead it’s equal to $-\sum_{n=q+1}^p a_n$ . And actually, I’ve made an off-by-one error: $\sum_{n=p}^q a_n$ should actually be $S_q - S_{p-1}$ . So this exercise is perhaps not as simple we thought! The first challenge was to figure out the “big idea” or strategy to use (i.e., recognize that Cauchyness is relevant), and the second challenge is now to make our idea actually work by getting all of the details right. Let’s write out all the variations of the claims we are working with, to get things straight:

$(S_N)_{N=m}^\infty$ is Cauchy (a)

For all $\varepsilon > 0$ there exists $N \geq m$ such that for all $p,q \geq N$ we have $|S_q - S_p| \leq \varepsilon$ (b)

For all $\varepsilon > 0$ there exists $N \geq m$ such that for all $p,q \geq N$ we have $|S_q - S_{p-1}| \leq \varepsilon$ (c)

For all $\varepsilon > 0$ there exists $N \geq m$ such that for all $p,q \geq N$ we have $\left|\sum_{n=p}^q a_n\right| \leq \varepsilon$ (d)

Our goal is to show that (a) and (d) are the same (why do we want to show both directions? It’s because the original Proposition is stated as an “if and only if”). That will allow us to use the fact that a sequence is Cauchy iff it is convergent.

It’s clear that (a) and (b) are equivalent, since (b) is just what it means for a sequence to be Cauchy. So we just need to show that (b) and (c) are equivalent and that (c) and (d) are equivalent.

(c) implies (b) since if $|S_q - S_{p-1}| \leq \varepsilon$ holds for all $p,q \geq N$ , then since $p+1 \geq N$ as well, we have $|S_q - S_{(p+1)-1}| \leq \varepsilon$ , i.e., $|S_q - S_p| \leq \varepsilon$ .

(b) implies (c) since if we have some $N$ such that $|S_q - S_p| \leq \varepsilon$ for all $p,q \geq N$ , then we can pick $N+1$ and see that if $p,q \geq N+1$ then $p-1,q \geq N$ and so $|S_q - S_{p-1}| \leq \varepsilon$ .

(The above two paragraphs are similar to Exercise 6.1.3 and Exercise 6.1.4. In all cases, what matters is what happens at the limits so shifting things around a bit won’t change whether a statement is true.)

So now we just have to show that (c) and (d) are equivalent. Being very mindful of off-by-one errors now, if $q \geq p$ we have $\sum_{n=p}^q a_n = S_q - S_{p-1}$ , and if $q < p$ instead we have $\sum_{n=p}^q a_n = 0$ and

$\begin{aligned}S_q - S_{p-1} &= (a_1 + \cdots + a_q) - (a_1 + \cdots + a_q + a_{q+1} + \cdots + a_p) \\ &= -(a_{q+1} + \cdots + a_p) \\ &= -\sum_{n=q+1}^p a_n\end{aligned}$

So whatever $p,q$ are, we have $\left|\sum_{n=p}^q a_n\right|\leq |S_q - S_{p-1}|$ . This means (c) implies (d): if the bigger thing is less than $\varepsilon$ , then so is the smaller thing. To show that (d) implies (c) as well, let $\varepsilon > 0$ . Since (d) is true, we have some $N$ ; pick this $N$ . Now let $p,q \geq N$ . If $q \geq p$ , we have $\sum_{n=p}^q a_n = S_q - S_{p-1}$ so we have $|S_q - S_{p-1}| = \left|\sum_{n=p}^q a_n\right| \leq \varepsilon$ . If $q < p$ then from (d) we have $\left|\sum_{n=q}^p a_n\right| \leq \varepsilon$ . But we know that $S_q - S_{p-1} = -\sum_{n=q+1}^p a_n$ , so actually we want to show that $\left|\sum_{n=q+1}^p a_n\right| \leq \varepsilon$ . But this is possible since $q+1 \geq N$ . So in other words we have

$\displaystyle |S_q - S_{p-1}| = \left|\sum_{n=q+1}^p a_n\right| \leq \varepsilon$

That was a lot of fiddly details, but we’re finally done.

Model solution

Let $S_N = \sum_{n=m}^N a_n$ be the $N$ th partial sum of the series $\sum_{n=m}^\infty a_n$ . By definition of series convergence, $\sum_{n=m}^\infty a_n$ converges iff the sequence $(S_N)_{N=m}^\infty$ converges.

We will show that $(S_N)_{N=m}^\infty$ is Cauchy if and only if, for every $\varepsilon > 0$ there exists an integer $N \geq m$ such that for all $p,q \geq N$ we have $\left|\sum_{n=p}^q a_n \right| \leq \varepsilon$ . If we can do this, then Theorem 6.4.18 will guarantee the result that we are trying to show.

First suppose that $(S_N)_{N=m}^\infty$ is Cauchy. Let $\varepsilon > 0$ . Since $(S_N)_{N=m}^\infty$ is Cauchy, this means that there exists an integer $N \geq m$ such that for all $p,q \geq N$ we have $|S_q - S_p| \leq \varepsilon$ . Consider the number $N+1$ . If $p,q \geq N+1$ , then $q, p-1 \geq N$ . Now we have two cases. If $q \geq p$ , then the sum $\sum_{n=p}^q a_n$ is equal to $S_q - S_{p-1}$ (Lemma 7.1.4(a)), so we have $\left|\sum_{n=p}^q a_n\right| = |S_q - S_{p-1}| \leq \varepsilon$ by Cauchyness. If on the other hand $q < p$ , then $\left|\sum_{n=p}^q a_n\right| = 0 < \varepsilon$ (Definition 7.1.1). Let us summarize: we let $\varepsilon > 0$ , then found a number $N+1 \geq m$ such that if $p,q \geq N+1$ then $\left|\sum_{n=p}^q a_n\right| \leq \varepsilon$ . So we’ve completed the first direction of the proof.

Now suppose that for every $\varepsilon > 0$ there exists an integer $N \geq m$ such that for all $p,q \geq N$ we have $\left|\sum_{n=p}^q a_n \right| \leq \varepsilon$ . Let $\varepsilon > 0$ . From our assumption, we are given some $N \geq m$ , so pick this $N$ . Now let $p,q \geq N$ . We have $\left|\sum_{n=p}^q a_n \right| \leq \varepsilon$ , and we want to show $|S_q - S_p| \leq \varepsilon$ . We have three cases:

If $q=p$ then $|S_q - S_p| = 0 < \varepsilon$ .
Next, if $q > p$ then $q \geq p+1$ and we have $\sum_{n=p+1}^q a_n = S_q - S_p$ by Lemma 7.1.4(a). And since $q,p+1 \geq N$ we can say that $\left|\sum_{n=p+1}^q a_n \right| \leq \varepsilon$ . But this means $|S_q - S_p| \leq \varepsilon$ as required.
Finally, if $q < p$ then $q+1 \leq p$ and we have $S_q - S_p = -(S_p - S_q) = -\sum_{n=q+1}^p a_n$ . But since $q+1, p \geq N$ we have $\left|\sum_{n=q+1}^p a_n\right| \leq \varepsilon$ , so $|S_q - S_p| = |S_p - S_q| = \left|\sum_{n=q+1}^p a_n\right| \leq \varepsilon$ as required.

In all three cases we have shown that $|S_q - S_p| \leq \varepsilon$ so we are done.

Exercise statement

Prove Proposition 6.6.5. (Note that one of the two implications has a very short proof.)

Proposition 6.6.5 (Subsequences related to limits). Let $(a_n)_{n=0}^\infty$ be a sequence of real numbers, and let $L$ be a real number. Then the following two statements are logically equivalent (each one implies the other):

(a) The sequence $(a_n)_{n=0}^\infty$ converges to $L$ .
(b) Every subsequence of $(a_n)_{n=0}^\infty$ converges to $L$ .

Hints

For one of the directions, use Lemma 6.6.4.

How to think about the exercise

This is a pretty straightforward exercise. I could leave it at that, but I want to try something new. So here’s a kind of stream-of-consciousness account of how I approached the problem:

We want to show (a) and (b) are the same. So first suppose (a) is true. We want to show (b) is true. What does that mean? Let’s unwrap:

Let $(a_n)_{n=0}^\infty$ be a sequence of real numbers, and suppose $(a_n)_{n=0}^\infty$ converges to $L \in \mathbf R$ . Now we want to show that every subsequence of $(a_n)_{n=0}^\infty$ converges to $L$ . It would be good to have an arbitrary subsequence and to give it a name. So let $(b_n)_{n=0}^\infty$ be a subsequence of $(a_n)_{n=0}^\infty$ . What does this mean? By definition, this means that there is a strictly increasing function $f : \mathbf N \to \mathbf N$ such that $b_n = a_{f(n)}$ for all $n \in \mathbf N$ .

Now, we want to show that $(b_n)_{n=0}^\infty$ converges to $L$ . How do we do that? To show that any sequence converges at all, the usual thing to do is to let $\varepsilon > 0$ be arbitrary. Now we have to find some $N$ such that $n \geq N$ implies $|b_n - L| \leq \varepsilon$ . How do we find such an $N$ ? The only possible way is to use some fact about $(a_n)_{n=0}^\infty$ , because that’s the only sequence we know anything about. We know that $(a_n)_{n=0}^\infty$ converges to $L$ . That means that for any $\varepsilon > 0$ (including the one we defined earlier), there is $N$ such that $n \geq N$ implies $|a_n - L| \leq \varepsilon$ . Okay, that is pretty close to what we want. Since we don’t know any other $N$ to use, let’s just pick this $N$ that we know exists, for now. We know that:

if $n \geq N$ then $|a_n - L|\leq \varepsilon$ . (*)

We want to show that:

if $n \geq N$ then $|b_n - L|\leq \varepsilon$ ;

or, making the relation between $(a_n)_{n=0}^\infty$ and $(b_n)_{n=0}^\infty$ clear, we want to show that:

if $n \geq N$ then $|a_{f(n)} - L|\leq \varepsilon$ .

So in other words, assuming $n \geq N$ , we have $|a_n - L|\leq \varepsilon$ and we want $|a_{f(n)} - L|\leq \varepsilon$ . Can we get what we want? We could if we knew that $f(n) \geq N$ (because of (*)). In other words, we want to know if $n \geq N$ implies $f(n) \geq N$ . One way this could be true is if $f(n) \geq n$ , so that $f(n) \geq n \geq N$ . (Here I have “cheated” a bit. I have seen the pattern “ $f(n) \geq n$ ” many times before, so it was immediately obvious to me that I should use it. I tried to give an account of how one might have reached the same conclusion without having known the pattern ahead of time, but I don’t know if I’ve succeeded.) This seems pretty plausibly true. The identity function seems like the slowest-growing strictly increasing function (it just counts up one by one), but other strictly increasing functions can be faster-growing (they can skip some numbers). It seems like an easy case of induction. It’s clear that $f(0) \geq 0$ , and if $f(n) \geq n$ then since $f(n+1) > f(n)$ , we have $f(n+1) > n$ , so $f(n+1) \geq n+1$ . So that completes the proof that (a) implies (b).

Now we do the other direction. Suppose every subsequence of $(a_n)_{n=0}^\infty$ converges to $L$ . We want to show that $(a_n)_{n=0}^\infty$ converges to $L$ . Since we know something about every subsequence, it seems like we ought to pick some clever particular subsequence to get what we want – oh, we can just pick $(a_n)_{n=0}^\infty$ to be the subsequence!

Model solution

First we show that (a) implies (b). Suppose the sequence $(a_n)_{n=0}^\infty$ converges to $L$ . We want to show that every subsequence of $(a_n)_{n=0}^\infty$ converges to $L$ . So let $(b_n)_{n=0}^\infty$ be a subsequence of $(a_n)_{n=0}^\infty$ . This means there is some strictly increasing function $f : \mathbf N \to \mathbf N$ such that $b_n = a_{f(n)}$ for all $n \in \mathbf N$ . We will show that $(b_n)_{n=0}^\infty$ converges to $L$ . Let $\varepsilon > 0$ . Since $(a_n)_{n=0}^\infty$ converges to $L$ , there exists some $N \geq 0$ such that:

for all $n\geq N$ we have $|a_n - L| \leq \varepsilon$ . (*)

Pick this $N$ , and let $n \geq N$ . We want to show that $|b_n - L| \leq \varepsilon$ . If we can show that $f(n) \geq n$ for all $n \in \mathbf N$ , then we would have $f(n) \geq n \geq N$ , so by (*) we have $|a_{f(n)} - L| \leq \varepsilon$ . By definition of $(b_n)_{n=0}^\infty$ this means we have $|b_n - L| \leq \varepsilon$ as required.

So it remains to show that $f(n) \geq n$ for all $n \in \mathbf N$ . We proceed by induction. The base case $f(0) \geq 0$ follows since the range of $f$ is $\mathbf N$ . Now suppose inductively that $f(n) \geq n$ . Since $f$ is strictly increasing, we have $f(n+1) > f(n)$ . Thus we have $f(n+1) > n$ . By Proposition 2.2.12(e) we have $f(n+1) \geq n+1$ as required. This closes the induction.

Next we show that (b) implies (a). This follows immediately from Lemma 6.6.4: $(a_n)_{n=0}^\infty$ is a subsequence of itself, so if every subsequence of $(a_n)_{n=0}^\infty$ converges to $L$ then in particular $(a_n)_{n=0}^\infty$ must converge to $L$ .

Tao Analysis Solutions

Monthly Archives: May 2023

Exercise 7.2.2

Exercise statement

Hints

How to think about the exercise

Model solution

Exercise 6.6.4

Exercise statement

Hints

How to think about the exercise

Model solution