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(1) 求级数 $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\left(1+n^{\sin x}\right)\left(1+n^...
Apr 23, 2024
(1) 求级数 ∑n=1∞(−1)n(1+nsin⁡x)(1+ncos⁡x)\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\left(1+n^{\sin x}\right)\left(1+n^{\cos x}\right)} 的收敛域.
Solution by Steps
step 1
To determine the convergence of the series ∑n=1∞(−1)n(1+nsin⁡x)(1+ncos⁡x)\sum_{n=1}^{\infty} \frac{(-1)^{n}}{(1+n^{\sin x})(1+n^{\cos x})}, we will use the Alternating Series Test (Leibniz's theorem)
step 2
The Alternating Series Test states that an alternating series ∑n=1∞(−1)nan\sum_{n=1}^{\infty} (-1)^{n}a_n converges if the sequence ana_n is positive, decreasing, and approaches zero as nn goes to infinity
step 3
We need to show that an=1(1+nsin⁡x)(1+ncos⁡x)a_n = \frac{1}{(1+n^{\sin x})(1+n^{\cos x})} is positive, decreasing, and approaches zero as n→∞n \to \infty for the series to converge
step 4
Since n^{\sin x} > 0 and n^{\cos x} > 0 for all n≥1n \geq 1, it follows that ana_n is positive for all nn
step 5
To show that ana_n is decreasing, we must show that a_{n+1} < a_n for all nn. This is equivalent to showing that (1+(n+1)^{\sin x})(1+(n+1)^{\cos x}) > (1+n^{\sin x})(1+n^{\cos x}) for all nn
step 6
As n→∞n \to \infty, both nsin⁡xn^{\sin x} and ncos⁡xn^{\cos x} approach infinity, so ana_n approaches zero
step 7
Therefore, the series converges for all xx because the conditions of the Alternating Series Test are satisfied
Answer
The series ∑n=1∞(−1)n(1+nsin⁡x)(1+ncos⁡x)\sum_{n=1}^{\infty} \frac{(-1)^{n}}{(1+n^{\sin x})(1+n^{\cos x})} converges for all xx.
Key Concept
Alternating Series Test (Leibniz's theorem)
Explanation
The series in question is an alternating series, and it satisfies the conditions of the Alternating Series Test, which are that the terms must be positive, decreasing, and approach zero. Since these conditions are met for all xx, the series converges for all xx.
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