Question

1. 1.11/1.11 points practice another determine whether the sequence converges or diverges. if it converges, find the limit. (if the sequence diverges, enter diverges.) ( a_n = ln(3n^2 + 4) - ln(n^2 + 4) ) ( lim_{n \to infty} a_n = square )

Explanation:

Step1: Use the logarithmic property

Use the property \(\ln a-\ln b=\ln\frac{a}{b}\). So, \(a_{n}=\ln(3n^{2} + 4)-\ln(n + 4)=\ln\frac{3n^{2}+4}{n + 4}\).

Step2: Divide numerator and denominator by \(n\)

Divide numerator and denominator of \(\frac{3n^{2}+4}{n + 4}\) by \(n\): \(\frac{3n^{2}+4}{n + 4}=\frac{3n+\frac{4}{n}}{1+\frac{4}{n}}\).

Step3: Find the limit of the fraction inside the log

As \(n\to\infty\), \(\lim_{n
ightarrow\infty}\frac{3n+\frac{4}{n}}{1+\frac{4}{n}}\). Since \(\lim_{n
ightarrow\infty}\frac{4}{n}=0\), we have \(\lim_{n
ightarrow\infty}\frac{3n+\frac{4}{n}}{1+\frac{4}{n}}=\lim_{n
ightarrow\infty}\frac{3n}{1}=\infty\).

Step4: Find the limit of \(a_{n}\)

Since \(y = \ln x\) and \(\lim_{x
ightarrow\infty}\ln x=\infty\), and we found that \(\lim_{n
ightarrow\infty}\frac{3n^{2}+4}{n + 4}=\infty\), then \(\lim_{n
ightarrow\infty}a_{n}=\lim_{n
ightarrow\infty}\ln\frac{3n^{2}+4}{n + 4}=\infty\).

Answer:

DIVERGES