Answer
The sequence $c_n$ converges to $ \ln
\left(\frac{2 }{3 }\right).$
Work Step by Step
We have
$$\lim_{n\to \infty}c_n=\lim_{n\to \infty}
\ln \left(\frac{2 n+1}{3 n+4}\right)
\\
=\ln \lim_{n\to \infty}
\left(\frac{2 n+1}{3 n+4}\right)=\ln \lim_{n\to \infty}
\left(\frac{2 +1/n}{3 +4/n}\right)\\
= \ln
\left(\frac{2 }{3 }\right).$$
Hence, by Theorem 1, the sequence $c_n$ converges to $ \ln
\left(\frac{2 }{3 }\right).$
