0
$\begingroup$

enter image description here

Hello. I'm having trouble understanding the answer to this question. It says that it is sufficient to show that the partial sums of $\sum{b_n}$ will form a subsequence of the partial sums of $\sum{a_n}$. That bit, is obvious. Also, removing the zero terms from the series will not change the sum which is clear as well. However, I do not get the answer to this (provided in the picture). Shouldn't every subsequence of the partial sums of $\sum{a_n}$ converge to A for $\sum{a_n}$ to be convergent to A? Why will one subsequence's convergence to A be sufficient to claim that the whole sequence and therefore $\sum{a_n}$ converges to A?

Can anyone please explain this?

Thanks in advance.

$\endgroup$
1
$\begingroup$

In general, you are correct about the fact that a sequence converges to $A$ if and only if all its subsequences also converge to $A$. In the light of this observation, it does not seem clear why if $ \sum b_n$, which is a subsequence of $\sum a_n$, converges then so does $ \sum a_n$.

However, if you fix a subsequence of $a_n$ which carries the same convergence structure(I will elaborate on this later) as $a_n$, then indeed the convergence of this subsequence can tell you a lot more about $a_n$ then what some arbitrary subsequence may be able to tell you.

Let us now discuss why the above proposition holds true. Set $c_n = \sum_{k=1}^n a_k$ and $d_n = \sum_{k=1}^n b_k$ as the partial sums. We want to show that $c_n$ converges if and only if $d_n$ does, and they go to the same limit.

If $c_n$ converges, then you showed that $d_n$ is a subsequence of $c_n$, so it will also converge to the same limit. This part is clear.

For the other part, we must understand what is the relationship between $c_n$ and $d_n$.

Key point : Since $b_n \neq 0$ for all $n$ (these are omitted), we see that $d_n \neq d_{n-1}$ for all $n$. In essence, $d_n$ is like $c_n$ but consecutive similar elements are combined.

That is, let's say that $c_n$ looks like this(all variables represent numbers): $$ a,b,c,c,e,e,e,f,f,a,g,v,v,v,v,t,t,... $$ then $d_n$ looks like: $$ a,b,c,e,f,a,g,v,t,... $$

In other words, $d_n$ is no arbitrary subsequence of $c_n$ : it has the following two key properties.

  • If $h = c_m$ for some $m$ then $h = d_l$ for some $l$ i.e. $d_l$ contains every element of $c_l$, while being its subsequence(That is, the set of elements of both sequences is the same).

  • $d_n$ preserves the order of terms in $c_n$ i.e. if $a,b$ are terms of $c_n$ and $b$ comes after $a$, then $b,a$ are also terms of $d_n$, and there too $b$ comes after $a$. These two properties tell you that $d_n$ is more than just a subsequence. It is like a compression of $c_n$, telling you pretty much what is important about it : the elements, and the order in which they appear are all preserved.

Now, we can attack the question. Suppose that $d_n \to A$. We want to show that $c_n$ also converges to $A$.

First, fix $\epsilon > 0$. We want to find some $N \in \mathbb N$ so that $n > N \implies |c_n - A| < \epsilon$.

We know, by convergence of $d_n$, that there is some $N_d \in \mathbb N$ so that $n > N_d \implies |d_n - A| < \epsilon$. Of course, $d_n$ is a subsequence of $c_n$, so $d_{N_d}$ appears somewhere in $c_n$, say at position $M$.

Then, if $m > M$, $c_m$ appears somewhere in $d_n$, but also after $N_d$! Hence, by what we know about $d_n$, we get $|c_{m} - A| < \epsilon$.

Thus, $c_n \to A$.

Again, note the salient features of $d_n$. It is no ordinary subsequence, preserving all distinct terms of $c_n$ with their order. This is what I will refer to as "convergence structure" : while I cannot give a comprehensive definition, the idea is that for a sequence to converge, one doesn't need to know all its terms, but rather what happens "after infinitely many terms" with the sequence i.e. the first few terms don't matter.

Any subsequence that manages (even in some convoluted fashion) to convey this information to us, which $d_n$ does : it tells us all the terms of $c_n$ and the order the come in, is easily capable of telling us whether the sequence itself converges or not.

$\endgroup$
  • $\begingroup$ By $d_{N_d}$, you mean the value of the sequence of partial sums at $n=N_d$? Which, is equal to $c_M$? Again, Thanks a lot!!! $\endgroup$ – A.Asad Apr 1 '18 at 9:20
  • $\begingroup$ Yes, that is right. You are welcome! $\endgroup$ – астон вілла олоф мэллбэрг Apr 1 '18 at 23:37
  • $\begingroup$ Can you please have a look at this as well? math.stackexchange.com/questions/2717457/… $\endgroup$ – A.Asad Apr 3 '18 at 17:04
  • 1
    $\begingroup$ www3.nd.edu/~apilking/Math10560/Lectures/Lecture%2027.pdf in here, I found the statement that you want to prove in slide 45 out of 53, the reminder estimation theorem(for the problem concerning alternating series). But you look at the conditions : we want $b_{n+1} < b_n$ for all $n$. If this condition is not satisfied, then in fact the statement given to you is false. This is not included in your question, and in fact in Bartle's notes as well, so it's a mistake. $\endgroup$ – астон вілла олоф мэллбэрг Apr 16 '18 at 0:26
  • 1
    $\begingroup$ You are welcome. The minute I saw something in the statement, I thought it unfit, after which I had gone on the search mission looking for the question. It's called the "alternating series estimate". $\endgroup$ – астон вілла олоф мэллбэрг Apr 20 '18 at 22:42
0
$\begingroup$

If a sequence is a Cauchy sequence and one of its subsequences converges then the whole sequence converges to the exact same limit.

$\endgroup$
  • $\begingroup$ But that requires us to know that the sequence is convergent. Here, in the question, we don't know if the partial sums of $\sum{a_n}$ converge or not? Of course if we know that the sequence is cauchy, implying its convergent and then any subsequence will converge to the same limit! $\endgroup$ – A.Asad Mar 30 '18 at 12:08
  • $\begingroup$ no, often showing Cauchy properties is easier than showing convergence. Therefore, if you can show that your sequence has the Cauchy property the argument in your question is completed. $\endgroup$ – user412810 Mar 30 '18 at 12:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.