Multplication of series

Question

My textbook is taking about the Cauchy product and I don't quite understand it and it says that when multiplying series, the sum of the third one is equal to the product of the sums of first two series, but how the general term $a_{n}$ of that third series then looks like? For instance when you multiply $\sum_{n=1}^{\infty}x^n$ by it self, the sum of a "product series" is equal to the product of sums of each of the series that is in this case $\frac{1}{1-x} * \frac{1}{1-x}$. What is and how to get the general therm of that third series? Is it $\sum_{n=1}^{\infty}x^{2n}$?

Answer 1

In general, $$\left(\sum_{n\ge 0}a_nx^n\right)\left(\sum_{n\ge 0}b_nx^n\right)=\sum_{n\ge 0}\left(\sum^{n}_{k=0}a_kb_{n-k}\right)x^n$$ This can be seen by writing the terms out explicitly. \begin{align} &(a_o+a_1x+a_2x^2+a_3x^3+\cdots)(b_o+b_1x+b_2x^2+b_3x^3+\cdots)\\ &=(a_0+b_0)+(a_0b_1+a_1b_0)x+(a_0b_2+a_1b_1+a_2b_0)x^2+\cdots \end{align} For $\dfrac{1}{(1-x)^2}$, \begin{align} \left(\sum_{n \ge 0}x^n\right)\left(\sum_{n \ge 0}x^n\right) &=\sum_{n \ge 0}\left(\sum^n_{k = 0}(1)(1)\right)x^n\\ &=\sum_{n \ge 0}(n+1)x^n\\ &=1+2x+3x^2+4x^3+\cdots \end{align} This can be verified using the binomial theorem \begin{align} (1-x)^{-2} &=\sum_{n\ge 0}\binom{2+n-1}{n}x^n\\ &=\sum_{n \ge 0}(n+1)x^n\\ \end{align} since $\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}$

Answer 2

To understand this it is best to recall the situation for finite sums, or even better polynomials where this is quite intuitive.

If you have $\sum_{n=0}^k a_iX^i$ and $\sum_{n=0}^k b_i X^i$, then you can write the product as $\sum_{n=0}^{2k} c_iX^i$ with $c_i = \sum_{j = 0}^i a_{i-j}b_j$.

The Cauchy product of series $\sum_{i= 0}^{\infty} a_i$ and $\sum_{i= 0}^{\infty} b_i$ is by definition the series $\sum_{i= 0}^{\infty} c_i $ with $c_i = \sum_{j = 0}^i a_{i-j}b_j$.

In your textbook it likely says that if the series $\sum_{i= 0}^{\infty} a_i$ is absolutely convergent with limit $A$ and if the series $\sum_{i= 0}^{\infty} b_i$ is absolutely convergent with limit $B$, then the series $\sum_{i= 0}^{\infty} c_i $ where $c_i = \sum_{j = 0}^i a_{i-j}b_j$ is absolutely convergent with limit $AB$.

You can define the Cauchy product of two series for any two series. It is however not obvious that this series $\sum_{i= 0}^{\infty} c_i $ is convergent and also not that it has as limit the product of the two limits. This is the content of the result.

Added: Since you also tagged taylor-expansion let me stress that what I write above is for general series but of course applies to power series (within the smaller radius of convergence of the two series).