Given $n \in \mathbb{N}$, there exists an algebra $A_n$ such that $\text{Findim}(A_n) = n - 1$ but $r.\text{gl.dim}(A_n) = \infty$.

Consider the following quiver algebra $A_n$:

[![Quiver diagram][1]][1]

with the relation that the composition of any two arrows is zero, i.e., $r^2 = 0$.

Clearly, $A_n$ is a monomial algebra. According to the paper [Finitistic dimensions of finite dimensional monomial algebras](https://pdf.sciencedirectassets.com/272332/1-s2.0-S0021869300X04152/1-s2.0-002186939190062D/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEID%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEaCXVzLWVhc3QtMSJHMEUCIBWPFsKGB6Gjp7EQULg4Jk5SNSvBZfPn9TWm5C9L36D9AiEA5NkAmcRZBZ5HIwAVHBOQdFq1UEjhNTkxqTQ%2BEAf10g8quwUI6f%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FARAFGgwwNTkwMDM1NDY4NjUiDKS%2BcDuCK5VvgQFFhSqPBUPG%2BCuROVRheLXKy1eezoifBeu3OK3esUPjQnmH8qXeckdQbrJQ%2BvfJQzB9H%2Fq%2BQYrwFuC1qKPPEkj2x%2Bhf6HHejBw33ke2AJQJr0lOeExcOv8Y2LHIDMtvOaX3jF%2FML9Maen8QamVCEb3fHh1GSosRVOf3BnkzFvJxiln%2FBy0905VtGjMYTYJKM7VdOp4TyjwDY131yJEKHJZsUQ9hLvbUwowgzvG6ONf6nR5NjTBsvBePmyUmzNGKsQqKZdHP2YFdmxOUWfj4y6yqPQOhATTc%2FNOfRpprdjzjzVVrK%2FYsA%2B5Au3AAaGu25iDnHZnR92HrlfXStjmINd7bNWTstthrEe0Yhq6jBP1mBzWLcr6JeIYJOOdHwJniBf5zhn%2BN45xuMRg2Q9Zlec4WpyaKBVFYX7qNhs4hoQ%2FtmrvGalLiTIHiqhsofT0NgKjjZfWlQWQgMG3DTYcpOyKzN%2F8oejO5E5Rs%2B8TXy6U7tq0aXChJOfU1qhcJsfSv8Ttohpfx%2BkZ5HvdiVcEcgXSao2GwIHbmHHvvsvdTenXUZ8pfjSn4jOqyFJK7eOZ%2BpEKjQ7%2Fh1j5PfPwHoC%2FDSrxcTPz8Ke63Kgf7%2B3WD5C1MiNeoEMjnEhXFCGjSs7clFsEL%2F3J%2Bn1FWDc4MwwWB9u0ITn4%2BQQTiSfYQC7dA86ZhbC0JtaA2C8EbhSBIxEB0poCgeF567DpjHk5k0M2BtRPooiW4ecSc9naIDIcNDxNdF8XcJCvgbwCRQUR5l1AvoPCoNGnyFn%2BVSUDsIFazrkKXDncQQd70kJTHqgdtuV7zG4pLUSEwX%2BWlpoqRFoNrVrE4zCtB66JBgUhjVbhyCJH9fgQtiCPZcHbbkrnFwrgm1xq7VtMw05ztuAY6sQEkce7oWeozBmkWPRCfOzFnFQBGvnYHi4kRxY0ekEektWVApX6Zdw5qhPkC9CjRE7SMDCRLm5oo4xwNi2XFWrKT%2Fy%2BOdBd2PnBO6blFNwW%2B%2FqrnjeVFdcj4iINpncEtT7kIOEcefhHdujGWspBxP3SzdD%2BbO%2FIyhIH2R%2BEe4atUW642CawuUrUfbtfNAA9iL4a6l9f8at4Pk8HrjsGl8obYzMh3KPnOFLw1hFNw4M0ZZwo%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20241025T085022Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYXGJ44JNQ%2F20241025%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=0020eb0aea55e4f01297973ba2a0cd29c5107cb6140883d36f79b4f3a115af96&hash=c96838070de6d475e8c1a91ac575cec2ead81757a72e2e6c5da174ea39b7e9ee&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=002186939190062D&tid=spdf-3d70a0be-47a9-4501-8797-b8b2216d30a5&sid=9eeb0dcb40aea540ce39207839d18c9ae3c7gxrqa&type=client&tsoh=d3d3LnNjaWVuY2VkaXJlY3QuY29t&ua=190b5f0b59500250580205&rr=8d80f56bde90dd5f&cc=cn), it follows that $\text{Findim}(A_n) = n-1$. Moreover, it is straightforward to compute that $r.\text{gl.dim}(A_n) = \infty$.

Therefore, we have $\text{Findim}(A_n) = n-1 < \infty = r.\text{gl.dim}(A_n)$ for $n \geq 1$.

Specifically, when $n=1$, $A_1 \cong k[t]/(t^2)$, so $\text{Findim}(k[t]/(t^2)) = \text{Findim}(A_1) = 1-1 = 0$.

[1]: https://i.sstatic.net/843h8YTK.png