Are all IIR filters unstable in nature?
Note that for stable IIR filters, the impulse response does approach zero as $n$ goes to infinity. It just never becomes exactly zero. However, the sum of the absolute values is finite. Just as an example, take the exponential impulse response

$$h[n]=a^nu[n],\qquad |a|&lt;1\tag{1}$$

where $u[n]$ is the unit step function. The sum

$$\sum_{n=-\infty}^{\infty}|h[n]|=\sum_{n=0}^{\infty}|a|^n=\frac{1}{1-|a|}\tag{2}$$

is obviously finite (for $|a|&lt;1$).

I agree with Hilmar's answer, but I would like to add that everything he said about poles is only valid for causal filters. In the general case it is not necessary for a stable filter to have all its poles inside the unit circle. For stability it is necessary and sufficient that the unit circle is inside the region of convergence (ROC) of the filter's $\mathcal{Z}$-transform (transfer function). So a stable filter can have poles inside as well as outside the unit circle, but in that case it can't be causal. So it's only the additional requirement of causality that forces the poles of a stable IIR filter inside the unit circle.

A similar thing holds for FIR filters. Causal FIR filters indeed have all their poles at the origin, but non-causal FIR filters also have poles at infinity. A simple example is this non-causal moving average filter

$$H(z)=\frac15\left(z^{-2}+z^{-1}+1+z+z^2\right)\tag{3}$$

which has poles at $z=0$ as well as poles at $z=\infty$, but being an FIR filter, it is obviously stable.