Examples of problems that are easier in the infinite case than in the finite case.

One can compute the value of $$\int_{0}^{\infty}e^{-t^{2}}\,\mathrm{d}t$$ exactly. This is known as the Gaussian integral, and it has its own Wikipedia page. The answer turns out to be $\frac{1}{2}\sqrt{\pi}.$

But one cannot do the same with $$\int_{0}^{x}e^{-t^{2}}\,\mathrm{d}t$$ because the antiderivative of the integrand is not an elementary function. This is why we gave a name to the error function $$\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^{2}}\,\mathrm{d}t,$$ which also has its own Wikipedia page.

In that sense, the infinite case is easier than the finite case.

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Addendum: The same phenomenon occurs for variants of this integral, in particular we can transform the integrand to evaluate $$\int_{-\infty}^{\infty}ae^{-(t-b)^{2}/(2c^{2})}\,\mathrm{d}t = \sqrt{2}a\lvert c\rvert \sqrt{\pi}$$ as detailed here on Wikipedia.

First thought would be series vs. partial sums (or improper vs. definite integrals) e.g.

$$ \sum_{k=0}^{\infty} \frac{1}{k!} \;=\; e $$

$$ \sum_{k=0}^{n} \frac{1}{k!} \;=\; \frac{e \cdot \Gamma(n+1,1)}{n!} $$

$\min c^Tx$

subject to $x \in P$

where $P$ is a bounded polyhedral can be solved easily by linear programming method. Ellipsoid method shows that the problem is in $P$ class.

$\min c^Tx$

subject to $x \in P \cap \mathbb{Z}^d$

that is imposing integer constraint increase the difficulty of the question.

Here $P$ is an infinite set in the sense that it contains infinitely many points but $P \cap \mathbb{Z}^d$ contains only finitely many points. Linear programming problem is easier than integer programming problem.