Which series converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. sum_{n=0}^infty(ln(4e^n-1)-ln(2e^n+1))

Question
Series
asked 2021-02-06
Which series converge, and which diverge? Give reasons for your answers. If a series converges, find its sum.
\(\sum_{n=0}^\infty(\ln(4e^n-1)-\ln(2e^n+1))\)

Answers (1)

2021-02-07
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.
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