Are there decidable problems for which for no algorithm we can give time bounds? Are there decidable problems such that for no algorithm which solves the problem we can give a time bound as a function of the length n of the input instance?

I arrived at this question because I was thinking about the following:

Assume we
have a recursively enumerable, but undecidable problem. Assume further that I is
a "yes"-instance of the problem. Then for no algorithm which identifies the
"yes"-instances of the problem we can give a time bound in terms of the size n
of I. For if we could give such a time bound, we could decide the problem, as we
could simply conclude that I is a "no"-instance when the time bound is exceeded.

Since we cannot give a time bound for recursively enumerable, undecidable
problems (for the computation time for "yes"-instances), I was wondering if there are decidable problems as well for which we
cannot give a time bound.
For every algorithm $A$ that terminates on a class of inputs $\mathsf{In}$, we can define the function of its running times:
$$
f(n)=\max_{i\in \mathsf{In(n)}}~~ \mathsf{time}(A(i)),
$$
where $\mathsf{In}(n)$ is the class of inputs of length $n$ and $\mathsf{time}(A(i))$ is the time algorithm $A$ requires to terminate on $i$. Of course, this definition is unsatisfying as it relates to the algorithm, but it shows the existence of such a function. 
The question that remains is whether there exists a concise representation (and I believe that this was what you were asking for).

If we use simple algebraic terms (without recursion of any sort) as definition of concise, then I think the answer is no: There are problems which can be decided but whose complexity is nonelementary. That is, there does not exist a stack of the form $2^{2^{2^{2^{\dots^n}}}}$ that bounds the execution time of an algorithm for a problem of size n. 

I hope I understood your question in the right way.