Question
\( S_{1}=\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}} \) \( S_{2}=\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) Which series converges?
Ask by Mejia Barber. in the United States
Feb 20,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
S₁ converges, and S₂ diverges.
Solution
We consider the series one by one.
For S₂:
S₂ = ∑ₙ₌₁∞ 1/√n is a p-series with p = 1/2, which is less than 1. A p-series converges only if p > 1. Therefore, the series S₂ diverges.
For S₁:
S₁ = ∑ₙ₌₁∞ (-1)ⁿ/√n is an alternating series. The Alternating Series Test tells us that if the absolute value of the terms decreases monotonically to 0, then the alternating series converges. In this case, the absolute value of the terms is 1/√n, which is indeed monotonically decreasing and tends to 0 as n → ∞. Thus, S₁ converges by the Alternating Series Test.
Conclusion:
S₁ converges (conditionally), while S₂ diverges.
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Simplify this solution Beyond the Answer
The series \( S_{1}=\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}} \) converges by the Alternating Series Test. This test states that if the absolute value of the terms \( a_n = \frac{1}{\sqrt{n}} \) is decreasing and approaches zero, the series converges. Here, \( a_n \) is indeed decreasing and approaches zero as \( n \) increases. On the other hand, the series \( S_{2}=\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) diverges. This can be shown using the p-series test, where \( p = \frac{1}{2} < 1 \). Since the p-series diverges for \( p \leq 1 \), \( S_{2} \) diverges. Therefore, only \( S_{1} \) converges.
