Q24·Section A (Section A)·24 of 60Convergence of two series with exponential decayLet an=e−2nsinna_n=e^{-2n}\sin nan=e−2nsinn and bn=e−nn2(sinn)2b_n=e^{-n}n^2(\sin n)^2bn=e−nn2(sinn)2 for n≥1n\ge1n≥1. Then OptionsA∑n=1∞an\sum_{n=1}^\infty a_n∑n=1∞an converges but ∑n=1∞bn\sum_{n=1}^\infty b_n∑n=1∞bn does NOT convergeB∑n=1∞bn\sum_{n=1}^\infty b_n∑n=1∞bn converges but ∑n=1∞an\sum_{n=1}^\infty a_n∑n=1∞an does NOT convergeCboth ∑n=1∞an\sum_{n=1}^\infty a_n∑n=1∞an and ∑n=1∞bn\sum_{n=1}^\infty b_n∑n=1∞bn convergeDNEITHER ∑n=1∞an\sum_{n=1}^\infty a_n∑n=1∞an NOR ∑n=1∞bn\sum_{n=1}^\infty b_n∑n=1∞bn convergesPick an option to check your answer.Show answerShow solution Report a problem with this question← Q23All questionsQ25 →Keyboard: A–D answer · S solution · ←/→ previous/next