Which series converge, and which diverge?If a series converges, find its sum.&nbsp;∑n=0∞(ln⁡(4en−1)−ln⁡(2en+1))

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Which series converge, and which diverge?If a series converges, find its sum.&amp;nbsp;∑n=0∞(ln⁡(4en−1)−ln⁡(2en+1))

i1ziZ · Accepted Answer

For the infinite series, locate the nth term.Infinite series:&amp;nbsp;∑n=0∞(ln⁡(4en−1)−ln⁡(2en+1))&amp;nbsp;So n-th term is a_n.&amp;nbsp;an=ln⁡(4en−1)−ln⁡(2en+1)&amp;nbsp;To determine whether the series converges or diverges, use the divergence test for the nth term.limn→∞an=limn→∞[ln⁡(4en−1)−ln⁡(2en+1)]&amp;nbsp;Use the properties for logarithm to simplify.&amp;nbsp;limn→∞an=limn→∞[ln⁡(4en−1)(2en+1)]&amp;nbsp;=limn→∞[ln⁡(4en−1)en(2en+1)en]&amp;nbsp;=limn→∞[ln⁡4−1en2+1en]&amp;nbsp;Further simplify.&amp;nbsp;limn→∞an=ln⁡(4−1e∞)(2+1e∞)&amp;nbsp;=ln⁡(4−1∞)(2+1∞)&amp;nbsp;=ln⁡(4−0)(2+0)&amp;nbsp;=ln⁡2≠0&amp;nbsp;So,&amp;nbsp;limn→∞an≠0&amp;nbsp;Hence, the series&amp;nbsp;∑n=0∞(ln⁡(4en−1)−ln⁡(2en+1))&amp;nbsp;diverges.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

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))

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