Let $A$ be a finite dimensional algebra over an algebraically closed field $\mathbb{K}$. Hence, $A$ can be realized as the path algebra of a bound quiver $(Q,I)$, where $I\subseteq\mathbb{K}Q$ is an admissible ideal.

Now, assume that $I\not=0$, and that $A$ is a triangular algebra, i.e., $Q$ doesn't have any oriented cycles. I'm interested in the case when $A$ is a string algebra. I'm trying to better understand the difference between $A$ being representation infinite and $A$ being $\tau$-tilting infinite when $A$ is a string algebra. So, if $A$ is representation finite, then $A$ is also $\tau$-tilting finite.

I'm wondering what is an example of a triangular string algebra $A$ that is representation infinite but $\tau$-tilting finite.

Note: I've edited the question because the previous version of the question wasn't specific enough. In the earlier version I didn't have the assumption that $A$ is a triangular algebra. Dave Benson's answer below is a perfectly valid example to my previous version of the question.

Let $A$ be a finite dimensional algebra over an algebraically closed field. I'm trying to better understand the difference between $A$ being representation infinite and $A$ being $\tau$-tilting infinite when $A$ is a string algebra. So, if $A$ is representation finite, then $A$ is also $\tau$-tilting finite.

I'm wondering what is an example of a string algebra $A$ that is representation infinite but $\tau$-tilting finite.

Let $A$ be a finite dimensional algebra over an algebraically closed field $\mathbb{K}$. Hence, $A$ can be realized as the path algebra of a bound quiver $(Q,I)$, where $I\subseteq\mathbb{K}Q$ is an admissible ideal.

Now, assume that $I\not=0$, and that $A$ is a triangular algebra, i.e., $Q$ doesn't have any oriented cycles. I'm interested in the case when $A$ is a string algebra. I'm trying to better understand the difference between $A$ being representation infinite and $A$ being $\tau$-tilting infinite when $A$ is a string algebra. So, if $A$ is representation finite, then $A$ is also $\tau$-tilting finite.

I'm wondering what is an example of a triangular string algebra $A$ that is representation infinite but $\tau$-tilting finite.

Note: I've edited the question because the previous version of the question wasn't specific enough. In the earlier version I didn't have the assumption that $A$ is a triangular algebra.

Let $A$ be a finite dimensional algebra over an algebraically closed field $\mathbb{K}$. Hence, $A$ can be realized as the path algebra of a bound quiver $(Q,I)$, where $I\subseteq\mathbb{K}Q$ is an admissible ideal.

Now, assume that $I\not=0$, and that $A$ is a triangular algebra, i.e., $Q$ doesn't have any oriented cycles. I'm interested in the case when $A$ is a string algebra. I'm trying to better understand the difference between $A$ being representation infinite and $A$ being $\tau$-tilting infinite when $A$ is a string algebra. So, if $A$ is representation finite, then $A$ is also $\tau$-tilting finite.

I'm wondering what is an example of a triangular string algebra $A$ that is representation infinite but $\tau$-tilting finite.

Note: I've edited the question because the previous version of the question wasn't specific enough. In the earlier version I didn't have the assumption that $A$ is a triangular algebra. Dave Benson's answer below is a perfectly valid example to my previous version of the question.

Let $A$ be a finite dimensional algebra over an algebraically closed field $\mathbb{K}$. Hence, $A$ can be realized as the path algebra of a bound quiver $(Q,I)$, where $I\subseteq\mathbb{K}Q$ is an admissible ideal.

Now, assume that $I\not=0$, and that $A$ is a triangular algebra, i.e., $Q$ doesn't have any oriented cycles. I'm interested in the case when $A$ is a string algebra. I'm trying to better understand the difference between $A$ being representation infinite and $A$ being $\tau$-tilting infinite when $A$ is a string algebra. So, if $A$ is representation finite, then $A$ is also $\tau$-tilting finite.

I'm wondering what is an example of a triangular string algebra $A$ that is representation infinite but $\tau$-tilting finite.

Note: I've edited the question because the previous version of the question wasn't specific enough. In the earlier version I didn't have the assumption that $A$ is a triangular algebra. Dave Benson's answer below is a perfectly valid example to my previous version of the question.

Let $A$ be a finite dimensional algebra over an algebraically closed field $\mathbb{K}$. Hence, $A$ can be realized as the path algebra of a bound quiver $(Q,I)$, where $I\subseteq\mathbb{K}Q$ is an admissible ideal.

Now, assume that $I\not=0$, and that $A$ is a triangular algebra, i.e., $Q$ doesn't have any oriented cycles. I'm interested in the case when $A$ is a string algebra. I'm trying to better understand the difference between $A$ being representation infinite and $A$ being $\tau$-tilting infinite when $A$ is a string algebra. So, if $A$ is representation finite, then $A$ is also $\tau$-tilting finite.

I'm wondering what is an example of a triangular string algebra $A$ that is representation infinite but $\tau$-tilting finite.

Note: I've edited the question because the previous version of the question wasn't specific enough. In the earlier version I didn't have the assumption that $A$ is a triangular algebra.