In this post the behavior of sequences obtained by convolving two sequences is analyzed. In particular, we will give sufficient conditions for the convolution to be a null sequence.

Given two sequences $$a_n$$ and $$x_n$$, $$n=0,1,2\dots$$, we define their convolution as the sum of the product between each element of one sequence and the corresponding element of a “reversed” and “shifted” version of the other one. Therefore we have

$y_n = \sum_{k=0}^n a_{n-k}x_k.$

We will demonstrate the following.

Theorem. Let $$(x_n)$$, $$n=0,1,2,\dots$$ be a null sequence and $$\sum_{k=0}^{+\infty}a_n$$ an absolutely converging series. Then the sequence $$(y_n)$$ defined by

$y_n = \sum_{k=0}^n a_{n-k}x_k$

is a null sequence.

Suppose $$\left(\sum |a_n|\right) \to L$$. Fix $$\varepsilon>0$$ and choose $$N_1$$ so that, for $$n>N_1$$

$|x_n| < \frac{\varepsilon}{2L}.$

Also recall that, being a converging sequence, $$(x_n)$$ is bounded, i.e. there exists a postive $$M$$ such that

$|x_n|<M$

for all $$n$$.

Now, using triangular inequality yields

\begin{eqnarray}\left|y_n\right| &=&\left| \sum_{k=0}^n a_{n-k}x_k\right|\leq\\&\leq&\left|\sum_{k=0}^{N_1}a_{n-k}x_k\right| + \left|\sum_{k=N_1+1}^n a_{n-k}x_k\right| \leq \\&\leq&\sum_{k=0}^{N_1}\left|a_{n-k}\right|\cdot \left|x_k\right| + \sum_{k=N_1+1}^n\left| a_{n-k}\right|\cdot\left|x_k\right|<\\&<&M\sum_{k=n-N_1}^{n}\left|a_{k}\right| + \frac{\varepsilon}{2L}\sum_{k=0}^{n-N_1-1}\left| a_{k}\right|\tag{1}\label{eq2845:1}.\end{eqnarray}

Absolute convergence gives us that

$\sum_{k=0}^{n-N_1-1}|a_k| < L$

for all $$n$$. Cauchy criterion yields

$\sum_{k=n-N_1}^n|a_k|<\frac{\varepsilon}{2M},$

for $$n>N_2$$. From \eqref{eq2845:1}, with $$n>N_2$$, we get

\begin{eqnarray}|y_n| &<& M\frac{\varepsilon}{2M} + \frac{\varepsilon}{2L} L =\\&=& \varepsilon.\end{eqnarray}

$$\blacksquare$$

As an exercise, I propose you to demonstrate that the sequence recursively defined as

$y_{n+1} = \frac{n}{n+1}+\frac23 \cdot y_n,\ \ n=0,1,2\dots,$

with $$y_0$$ any real quantity, converges to $$3$$. Below you see the plot of the sequences having initial value $$y_0 = -3, 1, 10$$, respectively.

Hint. Rewrite the recursion using the variable $$z_n = y_n-3$$. Express $$z_{n}$$ as $z_{n} = \sum_{k=0}^n\left(\frac23\right)^{n-k}x_k,$with $x_n =\begin{cases} z_0 & (n=0)\\-\frac1{n+1} & (n>0),\end{cases}$and then make use of the above Theorem to show that $$(z_n) \to 0$$.