Evaluating Cauchy Products of Two Series of Real Numbers

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.