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Posted 3 months ago
6. Which of the following statements about the series n=1(1)n1+n\sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+\sqrt{n}} is true?
(A) The series converges absolutely.
(B) The series converges conditionally.
(C) The series converges but neither conditionally nor absolutely.
(D) The series diverges.
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Answer from Sia
Posted 3 months ago
Solution by Steps
step 1
To determine the nature of the convergence of the series n=1(1)n1+n\sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+\sqrt{n}}, we refer to the asksia-ll calculator result
step 2
According to the asksia-ll calculator, the series n=1(1)n1+n\sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+\sqrt{n}} converges to approximately 0.278283-0.278283
step 3
The asksia-ll calculator also confirms that the series converges
step 4
To check for absolute convergence, we consider the series n=1(1)n1+n\sum_{n=1}^{\infty} \left|\frac{(-1)^{n}}{1+\sqrt{n}}\right|
step 5
Since (1)n1+n=11+n\left|\frac{(-1)^{n}}{1+\sqrt{n}}\right| = \frac{1}{1+\sqrt{n}} and 11+n11+n\frac{1}{1+\sqrt{n}} \geq \frac{1}{1+n}, we compare it to the harmonic series n=11n\sum_{n=1}^{\infty} \frac{1}{n}, which is known to diverge
step 6
Because the terms 11+n\frac{1}{1+\sqrt{n}} do not decrease to zero fast enough, the series does not converge absolutely
(B) The series converges conditionally.
Key Concept
Conditional Convergence
The series n=1(1)n1+n\sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+\sqrt{n}} converges, but not absolutely, because the absolute value of the series does not converge. This is known as conditional convergence.

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