\( S_{1}=\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}} \) \( S_{2}=\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) Which series converges?
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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.
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