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Say we have $\sum_{n=1}^{\infty} f(n,x)=f(x)$, which often happens with Taylor series:

Can we express:

$$\left(\sum_{n=1}^{\infty} f(n,x) \right)^2$$

As something that does not involve the square. I.e can multiply this out :

$$\left(f(1,x)+f(2,x)+f(4,x)+.... \right)\left(f(1,x)+f(2,x)+f(3,x)+...\right)$$

I know by distributive property we have:

$$=f(1,x)\sum_{n=1}^{\infty} f(n,x)+f(2,x)\sum_{n=1}^{\infty} f(n,x)+f(3,x)\sum_{n=1}^{\infty} f(n,x)+..$$

Can we simplify further?

Why I ask this is that I'm really interested if we can express

$\left(\frac{\ln (x)}{x-1} \right)^2$ as a series by only using the Taylor series for $\ln x$. I know I can just write the coefficients and multiply them out , but the new coefficients to the new series do not form an easy pattern to right down.

Jyrki Lahtonen
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Ahmed S. Attaalla
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1 Answers1

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What you are referring to is called the Cauchy product. If you consider multiplying out term by term, you might expect to see something like

$$(a_1+a_2+a_3+\cdots)(b_1+b_2+b_3+\cdots)=a_1b_1+(a_1b_2+a_2b_1)+(a_1b_3+a_2b_2+a_3b_1)+\cdots$$

where we are generalizing the FOIL method in the only way that makes sense.

The question is $(1)$ whether the sum converges and $(2)$ if it converges to what we expect it to (that is, if the first converges to $A$ and the second to $B$, then the product converges to $AB$). There is no answer in general but there are a few theorems to keep in mind:

$(1)$: if one (or both) of the series converges absolutely (and the other converges), then the product converges to $AB$.

$(2)$ if all three series converge, then the product indeed converges to $AB$.

In any particular case, like the one you mentioned, it may be difficult to show convergence of the product since the terms do not generalize well. One possibility is to find a bound for the product (perhaps using partial fractions) and show that the bound goes to zero face enough to guarantee convergence.

pancini
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  • What do you mean mean by if all three series converge? – Ahmed S. Attaalla Jul 11 '16 at 20:10
  • In your case, knowing that the series for $\frac{\ln x}{x-1}$ converges and that the product of that series with itself converges. – pancini Jul 11 '16 at 20:12
  • But isn't the "product of that series with itself" the thing we are trying to prove converges in the first place. Sorry I'm a bit confused. – Ahmed S. Attaalla Jul 11 '16 at 20:14
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    In a sense, yes. It is a bit confusing, but we can't always take the product of two convergent series and have it converge. You need to prove somehow that it does converge, and then the theorem tells you that it converges to the right thing. For there record, this is usually taught after at least a semester of real analysis. – pancini Jul 11 '16 at 20:19