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From my readings on the wikipedia, I was able to gather that the product of two infinite series $\sum_{i=0}^{\infty} a_{i}$ and $\sum_{j=0}^{\infty} b_{j} $ is outlined by the Cauchy Product. The cauchy product formula is explicitly shown below, $$ \sum_{i=0}^{\infty} a_i \sum_{j=0}^{\infty} b_j = \sum_{i=0}^{\infty} \sum_{j=0}^{i} a_{j} b_{i-j}. $$ However, I've saw this one Youtube video where multiplying two infinite series does not follow the Cauchy Product.

Example 1: $$\sum_{n=0}^{\infty} H_{n}(x) \frac{t^n}{n!} \, \sum_{m=0}^{\infty} H_{m}(x) \frac{s^m}{m!} = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{H_{n}(x) H_{m}(x)}{n! \, m!} t^n s^m $$

Picture 1: Involving the product of two generating functions. The youtube link to the video in Picture 1 is https://youtu.be/X7nlQFWv7bE?t=1m51s .

Is Example 1 valid and if so, why is it valid? Shouldn't the person have applied the Cauchy Product? Any resources that can point me in the right direction is appreciated.

og_kneecaps
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1 Answers1

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Quoting that very same Wikipedia article:

Convergence and Mertens' theorem

Let $(a_n)_n$ and $(b_n)_n$ be real or complex sequences. It was proved by Franz Mertens that, if the series $\sum_{n=0}^\infty a_n$ converges to $A$ and $\sum_{n=0}^\infty b_n$ converges to $B$, and at least one of them converges absolutely, then their Cauchy product converges to $AB$.

It is not sufficient for both series to be convergent; if both sequences are conditionally convergent, the Cauchy product does not have to converge towards the product of the two series [...]

(emphasis mine). The article then proceeds to show a counterexample in the case where the two series are only conditionally convergent.

In short: do not forget the assumptions.

Now, why is Example 1 valid? Well, you can definitely write, for all $N,M$, $$ \sum_{n=0}^N H_n(x) \frac{t^n}{n!} \cdot \sum_{m=0}^M H_m(x) \frac{s^m}{m!} = \sum_{n=0}^N\sum_{m=0}^M H_n(x)H_m(x) \frac{s^m}{m!}\frac{t^n}{n!} $$ Now, if both the left and the right expression converge when $N,M\to \infty$, you can get the identity. One typically has to prove it's the case, though.

Edit: a thing that may not be explicit: if you are asking whether it is possible to have two different expressions (Cauchy product and other) for the same product, well, yes. It is possible to have two expressions that look different, yet are equal.

Clement C.
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    If I know that $\sum_{n=0}^{\infty} H_{n}(x) \frac{t^n}{n!} $ converges and $\sum_{m=0}^{\infty}H_m(x) \frac{s^m}{m!}$ converges, can I say that these three statements are all equivalent as long as I know that each one of these statements converges? $$ \sum_{n=0}^{\infty} H_{n}(x) \frac{t^n}{n!} \sum_{m=0}^{\infty} H_{m}(x) \frac{s^m}{m!} = \sum_{n=0}^{\infty} \sum_{m=0}^{n} H_{m}(x) \frac{t^m}{m!} H_{n-m}(x) \frac{s^{n-m}}{(n-m)!} = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} H_{n}(x) \frac{t^m}{m!} H_{m}(x)\frac{s^m}{m!} $$ – og_kneecaps Jul 26 '16 at 03:20
  • Yes, this sounds right. The only statement to prove is that if all three series $\sum_n a_n$, $\sum_n b_n$, $\sum_n c_n$ (where $(c_n)_n$ is the Cauchy product of $(a_n)_n,(b_n)_n)$) converge, then indeed $$\sum_n a_n \sum_n b_n = \sum_n c_n.$$ This is true, e.g. as a consequence of Abel's theorem. – Clement C. Jul 26 '16 at 04:01
  • @ClementC. so am I right in thinking that if I know $\sum_1=\sum_{n=1}^\infty a_n=A$ converges and I know $\sum_2=\sum_{n=1}^\infty b_n=B$ converges and I can show that $\sum_3=\sum_{n=1}^\infty\sum_{m=1}^\infty a_nb_m$ converges (independently), then it must be true that $\sum_1\sum_2=\sum_3=AB$ ? Is there a name for this theorem? Or does it come from Abel's theorem you mention? – pshmath0 Nov 16 '19 at 15:28
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    @Pixel Then it seems like it's simply uniqueness of the limit. You have $A_n B_n = C_n$ for all $n$, and all three sequences $(A_n)_n, (B_n)_n, (C_n)_n$ converge. – Clement C. Nov 16 '19 at 19:25
  • @Pixel my bad, it's not as simple (namely, the "equality for all $n$" I claimed is... Not true. That's the point.) But what you say does indeed hold, and follow e.g., from Abel's theorem. (I.e., if the Cauchy product converges, it can only be to $AB$.) – Clement C. Nov 16 '19 at 22:38
  • @ClementC. Thanks. But just to clarify... when you say Cauchy product do you refer to $\sum_1\sum_2$ (which is what I intended) ? I don't want to consider $\sum_{n=1}^\infty\sum_{k=0}^n$ here, which is what I think of when one mentions Cauchy product. – pshmath0 Nov 16 '19 at 23:34
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    @Pixel Apologies, I misread. All I wrote above is for Cauchy products (i.e., what the OP and my answer were about). What you ask for is true, but pretty much by definition of the limit indeed. (I guess my first comment, which was wrong for Cauchy products, is ironically true for what you actually asked.) – Clement C. Nov 17 '19 at 00:29