It's a continuing task in that, after failure, the agent always gets a reward of 0 at each time-step ad infinitum.

From the book:

we could treat pole-balancing as a continuing task, using discounting. In this case the reward would be -1 on each failure and zero at all other times. The return at each time would then be related to -g^K, where K is the number of time steps before failure.

(Here I have used g as gamma, the discount factor).

Said another way, assuming the agent fails in the (K + 1)th step the reward is 0 till that step, -1 for it, and then 0 for eternity.

So the return: G_t = R_{t+1} + g R_{t+2} + g^2 R_{t+3} + ... + g^K R_{t+K+1} + ... = -g^K

It's a continuing task in that, after failure, the agent always gets a reward of $0$ at each time-step ad infinitum.

From the book:

we could treat pole-balancing as a continuing task, using discounting. In this case the reward would be -1 on each failure and zero at all other times. The return at each time would then be related to $-\gamma^K$, where $K$ is the number of time steps before failure.

(Here I have used $\gamma$ as the discount factor).

Said another way, assuming the agent fails in the (K + 1)th step the reward is $0$ till that step, $-1$ for it, and then $0$ for eternity.

So the return: $$G_t = R_{t+1} + \gamma R_{t+2} + \gamma^2 R_{t+3} + ... + \gamma^K R_{t+K+1} + ... = -\gamma^K$$