The Cauchy-product of these divergent, infinite series converges

Question

The sequences are defined as:

$a_n=\left\{\begin{array}{ll} 3, & n=0 \\ 3^n, & n\geq 1\end{array}\right. ,\quad b_n=\left\{\begin{array}{ll} -2, & n=0 \\ 2^n, & n\geq 1\end{array}\right. .$

$\sum \limits_{n=0}^{\infty}a_n,\quad\sum \limits_{n=0}^{\infty}b_n$

Assertion: The Cauchy-product of these infinite series converges

Proof:

$\sum \limits_{k=0}^{\infty}\sum \limits_{l=0}^{k}a_lb_{k-l}=a_0\sum \limits_{l=0}^{\infty}b_l+b_0\sum \limits_{l=1}^{\infty}a_l+\sum \limits_{k=1}^{\infty}\sum \limits_{l=1}^{k}a_lb_{k+1-l}=3\sum \limits_{l=0}^{\infty}2^l-2\sum \limits_{l=1}^{\infty}3^l+\sum \limits_{k=1}^{\infty}\sum \limits_{l=1}^{k}3^l\cdot2^{k+1-l}\\=\lim\limits_{l\to\infty}3\frac{1-2^{l+1}}{1-2}-\lim\limits_{l\to\infty}2\frac{1-3^{l+1}}{1-3}+\sum \limits_{k=1}^{\infty}2^{k+1}\sum \limits_{l=1}^{k}(\frac{3}{2})^l\\=\lim\limits_{l\to\infty}3\frac{1-2^{l+1}}{1-2}-\lim\limits_{l\to\infty}2\frac{1-3^{l+1}}{1-3}+\sum \limits_{k=1}^{\infty}2^{k+1}\lim\limits_{l\to\infty}\frac{1-(\frac{3}{2})^{l+1}}{1-(\frac{3}{2})}=-\infty$

I used the properties of a Geometric series.

I made a mistake, because it shouldn't diverge. Can somebody help me?

Answer 1

We calculate \begin{align} \sum_{k = 0}^\infty \sum_{l = 0}^k a_l b_{k - l} &= a_0 b_0 + a_0 b_1 + a_1 b_0 \\&+ \sum_{k = 2}^\infty \left[ a_0 b_k + a_k b_0 + \sum_{l = 1}^{k - 1} a_l b_{k - l} \right] \\&= 3 \cdot (-2) + 3 \cdot 2 + (-2) \cdot 3 \\&+ \sum_{k = 2}^\infty \left[ 3\cdot 2^k + (-2) \cdot 3^k + \sum_{l = 1}^{k - 1} 3^l \cdot 2^{k - l} \right] \\&= -6 + \sum_{k = 2}^\infty \left[ 3 \cdot 2^k + (-2) \cdot 3^k + 2^k \sum_{l = 1}^{k - 1} \left(\frac{3}{2}\right)^l \right] \\&= -6 + \sum_{k = 2}^\infty \left[ 3 \cdot 2^k + (-2)\cdot 3^k + 2^k \cdot \frac{3}{2} \cdot \frac{1 - \left(\frac{3}{2}\right)^{k-1}}{1 - \frac{3}{2}} \right] \\&= -6 + \sum_{k = 2}^\infty \left[3 \cdot 2^k - 2\cdot 3^k - 3\cdot 2^k \left(1 - \left(\frac{3}{2} \right)^{k-1} \right) \right] \\&= -6 + \sum_{k = 2}^\infty \left[ - 2\cdot 3^k + 3\cdot 2^k \left(\frac{3}{2} \right)^{k-1} \right] \\&= -6 + \sum_{k = 2}^\infty \left[ -2 \cdot 3^k + 2 \cdot 3^{k }\right] \\&= -6 + \sum_{k = 2}^\infty 0 = -6 \end{align}

Embarrassingly I did this calculation wrong the first time. If there's a moral there, it's that these sorts of calculations are easy to mess up and it doesn't hurt to check your work.

That said, the series does in fact converge.

Answer 2

The ordinary generating function of $\{a_n\}_{n\geq 0}$ is $A(x)=2+\frac{1}{1-3x}=3\cdot \frac{1-2x}{1-3x}$ and the ordinary generating function of $\{b_n\}_{n\geq 0}$ is $B(x)=-3+\frac{1}{1-2x}=2\cdot\frac{3x-1}{1-2x}$. In particular $A(x)B(x)=-6$ and $$ (a*b)_n = \sum_{k=0}^{n}a_k b_{n-k} $$ always equals $0$ unless $n=0$, where $(a*b)_0=-6$. In particular $\sum_{n\geq 0}(a*b)_n = -6$.

Answer 3

The Cauchy product does not necessarily converge! But this particular Cauchy product converges. Take the formula of the Cauchy product:

$c_{k}=\sum_{n=0}^{k}a_{n}b_{k-n}$.

Clearly $c_{0}=a_{0}b_{0}=-6$

Now

$\sum_{n=0}^{k}a_{n}b_{k-n}$=$a_{0}b_{k}+a_{1}b_{k-1}+a_{2}b_{k-2}+........+a_{k-1}b_{1}+a_{k}b_{0}$=

$=32^{k}+32^{k-1}+3^{2}2^{k-2}+......+3^{k-1}2+3^{k}(-2)$.

Put aside for a moment the first and the last term and focus on the rest of the terms.

We add and subtract $3^{k}+2^{k}$ and we get:

$(3^{k}+3^{k-1}2+3^{k-2}2^{2}+........+32^{k-1}+2^{k})-3^{k}-2^{k}$.

The sum in the parenthesis is $\dfrac{3^{k+1}-2^{k+1}}{3-2}=3^{k+1}-2^{k+1}$.

Now we sum together all the terms (including the first and the last that we have put aside) and we get:

$32^{k}+3^{k+1}-2^{k+1}-3^{k}-2^{k}+3^{k}(-2)$=$3^{k}(3-2)-3^{k}+2^{k}(3-2)-2^{k}=0$.

So $c_{0}=-6\,\,$ and all the rest of $\,\,c_{k}=0$.

Thus the Cauchy product converges despite the fact that $\sum_{n=0}^{+\infty}a_{n}$ and $\sum_{n=0}^{+\infty}b_{n}$ diverge!!