# Difficulty regarding understanding a proof of multiplication of infinite series

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.

• 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 – Jakobian Dec 27 '18 at 14:40
• @Jakobian Convergent series are always associative though... – Theo Bendit Dec 27 '18 at 14:55

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.

• yes, it is very much clear to me now – Arnab Chowdhury Dec 27 '18 at 16:12
• 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 – Arnab Chowdhury Dec 27 '18 at 16:14
• @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. – Noble Mushtak Dec 27 '18 at 16:44
• This theorem is basically Fubini's theorem for counting measure, isn't it. – Jakobian Dec 27 '18 at 16:48
• @Jakobian Yes, it is definitely very similar to Fubini's theorem. In fact, sometime's it is called Fubini's theorem for double series. – Noble Mushtak Dec 27 '18 at 18:16