I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 10.1 : Sequences
6. Determine if the given sequence converges or diverges. If it converges what is its limit?
\[\left\{ {\frac{{\ln \left( {n + 2} \right)}}{{\ln \left( {1 + 4n} \right)}}} \right\}_{n = 1}^\infty \]Show All Steps Hide All Steps
Start SolutionTo answer this all we need is the following limit of the sequence terms.
\[\mathop {\lim }\limits_{n \to \infty } \frac{{\ln \left( {n + 2} \right)}}{{\ln \left( {1 + 4n} \right)}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{}^{1}/{}_{{n + 2}}}}{{{}^{4}/{}_{{1 + 4n}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{1 + 4n}}{{4\left( {n + 2} \right)}} = 1\]You do recall how to use L’Hospital’s rule to compute limits at infinity right? If not, you should go back into the Calculus I material do some refreshing.
Show Step 2We can see that the limit of the terms existed and was a finite number and so we know that the sequence converges and its limit is one.