0
$\begingroup$

I am referring to article no $3.50$, from Principles of Mathematical Analysis by Walter Rudin. The theorem is

Let $\sum_{n=0}^\infty a_n=A$ and $\sum_{n=0}^\infty b_n=B$ be two convergent series. Their product is defined by $ c_n =\sum_{k=0}^n a_nb_{n-k}$. Then the series $\sum_{n=0}^\infty c_n$ converges to $AB$ if atleast one of the two series converges absolutely.

My approach:

$c_0=a_0b_0$ , $c_1=a_0b_1+a_1b_0$, $c_2=a_0b_2+a_1b_1+a_2b_0$ and so on...

$$\sum_{n=0}^\infty c_n= c_0+c_1+c_2+c_3+...$$ $$=a_0b_0+(a_0b_1+a_1b_0)+(a_0b_2+a_1b_1+a_2b_0)+(a_0b_3+a_1b_2+a_2b_1+a_3b_0)+...$$ $$=a_0(b_0+b_1+b_2+b_3+...)+a_1(b_0+b_1+b_2+b_3+...)+a_2(b_0+b_1+b_2+b_3+...)+...$$ $=a_0B+a_1B+a_2B+...=B(a_0+a_1+a_2+...)=AB$

There must be some flaw in my work as I need not to assume that at least one of them is absolutely convergent, please correct me where I went wrong. Thank you.

$\endgroup$
  • 3
    $\begingroup$ Infinite sums are not associative, and they're not commutative. You can't operate on them like on normal sums (most of the time). Try to prove it from definition $\endgroup$ – Jakobian Dec 27 '18 at 14:40
  • 1
    $\begingroup$ @Jakobian Convergent series are always associative though... $\endgroup$ – Theo Bendit Dec 27 '18 at 14:55
4
$\begingroup$

You have to be more careful with your rearrangements because rearranging terms in infinite series like this is not always valid. When in doubt, I always like to use switching the order of summations since there is a specific theorem that allows for that. However, in order to use this theorem, you first have to prove that the infinite sum of the series $d_n=\sum_{i=0}^n |a_ib_{n-i}|$ converges, which is where you would need to use the hypothesis that either $a_n$ or $b_n$ is absolutely convergent.

Anyway, the sum we are trying to find is the following: $$\sum_{n=0}^\infty c_n=\sum_{n=0}^\infty \sum_{i=0}^n a_ib_{n-i}$$

The second sum goes from $i=0$ to $n$, so $i \leq n$ at all times. However, another way to look at this is that $n \geq i$ at all times, so we can switch the order of summation and say that $n$ goes from $i$ to $\infty$.

$$\sum_{i=0}^\infty \sum_{n=i}^\infty a_ib_{n-i}$$

Luckily, we can factor out $a_i$ from the second summation since it has nothing to do with $n$:

$$\sum_{i=0}^\infty \left[a_i\sum_{n=i}^\infty b_{n-i}\right]$$

Now, we can change the index of the second summation by saying $z=n-i$, so it goes from $z=0$ to $\infty$ instead:

$$\sum_{i=0}^\infty \left[a_i\sum_{z=0}^\infty b_z\right]$$

Now, we can factor the second summation out from the first since it has nothing to do with $i$:

$$\left[\sum_{z=0}^\infty b_z\right]\left[\sum_{i=0}^\infty a_i\right]=BA$$

Hopefully, this gives you a better understanding of how to formally rearrange terms in infinite series using rearrangements and distributive property. The idea behind what I did is exactly the same as what you did with $a_0B+a_1B+...$, but this process is more formal and mathematically valid.

$\endgroup$
  • $\begingroup$ yes, it is very much clear to me now $\endgroup$ – Arnab Chowdhury Dec 27 '18 at 16:12
  • $\begingroup$ but can you please give me any reference related to the $d_n$ related theorem you mentioned earlier. I just want to get a bit more concrete concept. @Noble Mushtak $\endgroup$ – Arnab Chowdhury Dec 27 '18 at 16:14
  • $\begingroup$ @ArnabChowdhury Please see [this link](bit.ly/2RfzUff). Before you can switch order of summation, you have to prove $\sum_{n=0}^\infty\sum_{i=0}^n |a_nb_{n-i}|$ is a convergent sum. Notice that this is the same as the original sum, except that I have taken the absolute value of every term. Now, proving that this sum is convergent is the same as taking the series $d_n=\sum_{i=0}^n |a_nb_{n-i}|$ and then seeing if the infinite sum of that series converges. If you can prove $d_n$ has a convergent infinite sum, then that is the same as saying $\sum_{n=0}^\infty\sum_{i=0}^n |a_nb_{n-i}|$ converges. $\endgroup$ – Noble Mushtak Dec 27 '18 at 16:44
  • $\begingroup$ This theorem is basically Fubini's theorem for counting measure, isn't it. $\endgroup$ – Jakobian Dec 27 '18 at 16:48
  • $\begingroup$ @Jakobian Yes, it is definitely very similar to Fubini's theorem. In fact, sometime's it is called Fubini's theorem for double series. $\endgroup$ – Noble Mushtak Dec 27 '18 at 18:16

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.