Essentially, yes. The connection is this: if we take a sequence of partitions P_1, ..., P_n, ... of the interval, such that the norm of the partition ||P_n|| tends to 0 as n tends to infinity, and let S_1, ..., S_n, ... be the corresponding Riemann sums, then the Riemann integral converges if and only if all such sequences {S_n} converge. Or, in plainer language, we generate a value for the Riemann sum and successively tighten the partition; each time we tighten the partition we get a new value for the sequence.
If this sequence is guaranteed to converge no matter what specific partitions we're using, then the integral is said to converge. Note that it's possible for one specific such sequence to converge without the integral converging, if you have a pathological function and choose your partitions (and evaluation points) carefully. A good example is the Dirichlet function which is 0 for rational numbers and 1 for irrational numbers, say over the interval 0, 1.
If you only evaluate at rational numbers you will get a Riemann sum of 0; if you only evaluate at irrational numbers you will get a Riemann sum of 1. You can use either approach with an arbitrarily small partition, so you could get a constant sequence (0, 0, 0, ...) or (1, 1, 1, ...). But not all sequences of Riemann sums converge for this function; if you alternate between the two you will get (0, 1, 0, 1, ...) which does not converge.
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