Evaluating Cauchy Products of Two Series of Real Numbers

Recall from the The Cauchy Product of Two Series of Real Numbers page that if $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ are two series of real numbers then the Cauchy product of these series is defined to be the series $\displaystyle{\sum_{n=0}^{\infty} c_n}$ where:

(1)

\begin{align} \quad c_n = \sum_{k=0}^{\infty} a_kb_{n-k} \end{align}

On the Convergence of Cauchy Products of Two Series of Real Numbers page we saw that if $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ are absolutely convergent and converge to $A$ and $B$ respectively, then the corresponding Cauchy product converges absolutely to $AB$.

We even noted that if $\displaystyle{\sum_{n=0}^{\infty} a_n}$ is absolutely convergent and converges to $A$, while $\displaystyle{\sum_{n=0}^{\infty} b_n}$ is only conditionally convergent and converges to $B$, then the corresponding Cauchy product still converges (not necessarily absolutely though) to $AB$.

We will now look at some examples of evaluating Cauchy products.

Example 1

Show that the Cauchy product of two conditionally convergent series need not be convergent by showing that $\displaystyle{\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n+1}}}$ is conditionally convergent and that the Cauchy product of this series with itself is divergent.

Notice that $\displaystyle{\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n+1}}}$ is an alternating series, $\left ( \frac{1}{\sqrt{n+1}} \right )_{n=0}^{\infty}$ is decreasing, and $\lim_{n \to \infty} \frac{1}{\sqrt{n+1}} = 0$. So, by the The Alternating Series Test for Alternating Series of Real Numbers we have that $\displaystyle{\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n+1}}}$ converges.

Now let's determine $c_n$ for the Cauchy product of this series with itself. We have that:

(2)

\begin{align} \quad c_n = \sum_{k=0}^{n} \left ( \frac{(-1)^{k+1}}{\sqrt{k+1}} \right ) \left ( \frac{(-1)^{n - k + 1}}{\sqrt{n - k + 1}} \right ) = \sum_{k=0}^{n} \frac{(-1)^{n+2}}{\sqrt{k+1} \sqrt{n - k + 1}} = (-1)^{n+2} \sum_{k=0}^{n} \frac{1}{\sqrt{k+1} \sqrt{n-k+1}} \end{align}

Now notice that for all $n \in \mathbb{N}$ that:

(3)

\begin{align} \quad \frac{(-1)^{n+2}}{\sqrt{k+1} \sqrt{n - k + 1}} \geq \frac{(-1)^{n+2}}{\sqrt{n+1} \sqrt{n - k + 1}} \geq \frac{(-1)^{n+2}}{\sqrt{n+1} \sqrt{n + 1}} = \frac{(-1)^{n+2}}{n+1} \end{align}

So then we have that:

(4)

\begin{align} \quad \mid c_n \mid = \sum_{k=0}^{n} \frac{1}{\sqrt{k + 1}\sqrt{n - k + 1}} \geq \sum_{k=0}^{n} \frac{1}{n+1} = \frac{n+1}{n+1} = 1 \end{align}

So for all $n \in \mathbb{N}$ we have that $\mid c_n \mid \geq 1$ and so $\displaystyle{\lim_{n \to \infty} \mid c_n \mid \neq 0}$ which implies that $\displaystyle{\lim_{n \to \infty} c_n \neq 0}$, so the Cauchy product $\displaystyle{\sum_{n=0}^{\infty} c_n}$ diverges.

Example 2

Find the Cauchy product of the series $\displaystyle{\sum_{n=0}^{\infty} \frac{1}{n+1}}$ with itself. Does the Cauchy product converge?

For each $n \in \mathbb{N}$ we have that:

(5)

\begin{align} \quad c_n = \sum_{k=0}^{n} \frac{1}{k+1} \frac{1}{n - k + 1} \end{align}

Notice that:

(6)

\begin{align} \quad \frac{1}{k+1} \frac{1}{n - k + 1} \geq \frac{1}{n+1} \frac{1}{n+1} = \frac{1}{(n+1)^2} \end{align}

Therefore we see that:

(7)

\begin{align} \quad c_n = \sum_{k=0}^{n} \frac{1}{k+1} \frac{1}{n - k + 1} \geq \sum_{k=0}^{n} \frac{1}{(n+1)^2} = \frac{(n+1)}{(n+1)^2} = \frac{1}{n+1} \end{align}

We see that the series $\displaystyle{\sum_{n=0}^{\infty} \frac{1}{n+1}}$ diverges as a harmonic series and so by comparison, the Cauchy product $\displaystyle{\sum_{n=0}^{\infty} c_n}$ diverges.