Find the nth term for the infinite series.

Infinite series: \(\sum_{n=0}^\infty(\ln(4e^n-1)-\ln(2e^n+1))\)

So n-th term is a_n.

\(a_n=\ln(4e^n-1)-\ln(2e^n+1)\)

Use the divergence test for the nth term to find if the series converges or diverges.

\(\lim_{n\to\infty}a_n=\lim_{n\to\infty}[\ln(4e^n-1)-\ln(2e^n+1)]\)

Use the properties for logarithm to simplify.

\(\lim_{n\to\infty}a_n=\lim_{n\to\infty}[\ln\frac{(4e^n-1)}{(2e^n+1)}]\)

\(=\lim_{n\to\infty}[\ln\frac{\frac{(4e^n-1)}{e^n}}{\frac{(2e^n+1)}{e^n}}]\)

\(=\lim_{n\to\infty}[\ln\frac{4-\frac{1}{e^n}}{2+\frac{1}{e^n}}]\)

Further simplify. \(\lim_{n\to\infty}a_n=\ln\frac{(4-\frac{1}{e^\infty})}{(2+\frac{1}{e^\infty})}\)

\(=\ln\frac{(4-\frac{1}{\infty})}{(2+\frac{1}{\infty})}\)

\(=\ln\frac{(4-0)}{(2+0)}\)

\(=\ln2\ne0\)

So, \(\lim_{n\to\infty}a_n\ne0\)

Hence, the series \(\sum_{n=0}^\infty(\ln(4e^n-1)-\ln(2e^n+1))\) diverges.

Infinite series: \(\sum_{n=0}^\infty(\ln(4e^n-1)-\ln(2e^n+1))\)

So n-th term is a_n.

\(a_n=\ln(4e^n-1)-\ln(2e^n+1)\)

Use the divergence test for the nth term to find if the series converges or diverges.

\(\lim_{n\to\infty}a_n=\lim_{n\to\infty}[\ln(4e^n-1)-\ln(2e^n+1)]\)

Use the properties for logarithm to simplify.

\(\lim_{n\to\infty}a_n=\lim_{n\to\infty}[\ln\frac{(4e^n-1)}{(2e^n+1)}]\)

\(=\lim_{n\to\infty}[\ln\frac{\frac{(4e^n-1)}{e^n}}{\frac{(2e^n+1)}{e^n}}]\)

\(=\lim_{n\to\infty}[\ln\frac{4-\frac{1}{e^n}}{2+\frac{1}{e^n}}]\)

Further simplify. \(\lim_{n\to\infty}a_n=\ln\frac{(4-\frac{1}{e^\infty})}{(2+\frac{1}{e^\infty})}\)

\(=\ln\frac{(4-\frac{1}{\infty})}{(2+\frac{1}{\infty})}\)

\(=\ln\frac{(4-0)}{(2+0)}\)

\(=\ln2\ne0\)

So, \(\lim_{n\to\infty}a_n\ne0\)

Hence, the series \(\sum_{n=0}^\infty(\ln(4e^n-1)-\ln(2e^n+1))\) diverges.