Jtdylktuy

Does an infinite sequence always have finite entropy? For example, doesn't $a_n=n$, the sequence of non-negative integers, have very low entropy? It feels like all "well-defined" sequences ought to have low entropy because there's very little surprise about the next element.

I think it makes sense if the terms of your series are all positive and the series converges.
For example if $A=sum_{n=1}^infty a_n$, then you can define the probability distribution $p_n=a_n/A$. Hence, using the entropy definition $S=-sum_{n=1}^infty p_nln p_n$, which is the standard definition of statistical mechanics divided by the Boltzmann constant $k_B$. For example, we could define then the entropy of Riemann's zeta function $zeta(s)=sum_{n=1}^inftyfrac{1}{n^s}$. Using the probability distribution $p_n(s)=1/(zeta(s)n^s)$, we obtain the "entropy" of the
zeta function:
begin{align}
S(s) &= frac{1}{zeta(s)}sum_{n=1}^inftyfrac{1}{n^s}ln(zeta(s)n^s)=lnzeta(s)+frac{s}{zeta(s)}sum_{n=1}^inftyfrac{ln n}{n^s}=lnzeta(s)-frac{s}{zeta(s)}zeta'(s)\
&=lnzeta(s)-sfrac{dlnzeta(s)}{ds}
end{align}
This is called the zeta distribution.
See here:
https://en.wikipedia.org/wiki/Zeta_distribution
Another example is the Poisson distribution, which you can obtain from the
expansion of $e^{lambda t}$. The probability distribution you get from this
is $p_n=frac{e^{-lambda t}(lambda t)^n}{n!}$.

Your question is a little vague.

In the context of the Shannon entropy, one natural and usual measure of the "rate of information" of an infinite sequence (more aptly: of a discrete time stochastic process) is the entropy rate:

$$H_r = lim_{nto infty} frac{H(X_1,X_2 , cdots H_n)}{n}$$

... if the limit exists, of course. Notice also that this is not the information of a single full sequence, but a normalized expected value.

Typically, if $H_r >0$, then the entropy of the infinite sequence is also infinite.
In particular, if the sequence is produced by a stationary memoryless source (independent symbols with stationary distribution) then $H(X_1,X_2 , cdots H_n)=n H(X_1)$ and $H_r = H(X_1)$

A little more general: for a stationary first order Markov process, $H_r = H(X_2 mid X_1)$

If each symbol is totally predictable from the previous one, then $H_r=0$.

In your case, your sequence is not only predictable but also deterministic, hence $H_r=0$

This is not the end of the story, though. Because the Shannon entropy requires a probabilistic setting, and sometimes that does not seem very adequate. The typical example: which is the entropy rate of $X_n=$ digits of the decimal expansion of $pi$?
For an alternative approach to defining the "average information" of a sequence (and hence some alternative "entropy"), using an operational (computational) instead of probabilistic setting, you might look into Kolmogorov complexity