Home » Uncategorized » Limit of a sum

# Limit of a sum

Let be a real sequence such that , , and . Prove that

Solution

Lemma: Let be a bounded sequence of, say, complex numbers and let be another complex number.

where the latter means “convergence on a set of density ” i.e. there exists a subsequence such that   and is dense i.e.

Proof: The proof is omitted because it is too technical.

Hence,

All we need is that the logarithm is continuous and that as well as are bounded sequences.

Note: We can simplify the conditions to and .

The exercise along with the solution may be found on AoPS.com . The proof of the claim may also be found there.