Use the Ratio Test to determine whether the series is convergent or divergent.n=1∑∞(n+1)72n+114nIdentify an.Evaluate the following limit.n→∞lim∣∣anan+1∣∣Since limn→∞∣∣anan+1∣∣?∨1,−− Select-- ✓.
Q. Use the Ratio Test to determine whether the series is convergent or divergent.n=1∑∞(n+1)72n+114nIdentify an.Evaluate the following limit.n→∞lim∣∣anan+1∣∣Since limn→∞∣∣anan+1∣∣?∨1,−− Select-- ✓.
Identify General Term: Identify an as the general term of the series.an=(n+1)72n+114n
Evaluate Limit of Ratio: Evaluate the limit of the ratio of successive terms. \lim_{n \to \infty}\left|\frac{a_{n+\(1\)}}{a_{n}}\right| = \lim_{n \to \infty}\left|\frac{\left(\(14\)^{n+\(1\)}\right)/\left((n+\(2\))\(7\)^{\(2\)n+\(3\)}\right)}{\left(\(14\)^{n}\right)/\left((n+\(1\))\(7\)^{\(2\)n+\(1\)}\right)}\right|
Simplify Expression: Simplify the expression inside the limit.\(\newline=limn→∞∣∣(14n)(n+2)72n+3(14n+1)(n+1)72n+1∣∣=limn→∞∣∣72(n+2)14(n+1)∣∣=limn→∞∣∣49(n+2)14(n+1)∣∣
Further Simplify Limit: Further simplify the limit expression.=limn→∞∣∣7(n+2)2(n+1)∣∣
Evaluate Limit: Evaluate the limit as n approaches infinity.=limn→∞∣∣7(n+2)2(n+1)∣∣=72
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