91Ó°ÊÓ

When dealing with an infinite series in mathematics, it's crucial to understand when such a series is considered convergent. A series \( \sum_{n=1}^{\text{\infty}} a_n \) is said to converge if the sequence of its partial sums \( S_n = a_1 + a_2 + ... + a_n \) approaches a finite limit as \( n \) approaches infinity. In simpler terms, if you keep adding terms of the series indefinitely and the total sum stabilizes at a certain number, instead of growing without bounds, that series is convergent.

The series in our example \( \sum_{k=1}^{\text{\infty}} \frac{(-1)^{k}}{\sqrt{k}} \) is of particular interest because it is an alternating series, where the signs of the terms alternate between positive and negative. Despite this oscillation in signs, this series can still converge if specific criteria are met. Those criteria are enshrined in the Alternating Series Test, which requires that the absolute value of the terms must be both decreasing and approaching zero.

A sequence \( \{a_k\} \) is said to be decreasing if every term is less than or equal to the previous term, that is, \( a_{k+1} \leq a_k \) for all \( k \) in the domain of the sequence. For the series to pass the first condition of the Alternating Series Test, the absolute values of the terms must form a decreasing sequence.

In our step-by-step solution, we've determined that \( |a_k| = \frac{1}{\sqrt{k}} \) is indeed decreasing because \( |a_{k+1}| =\frac{1}{\sqrt{k+1}} \leq |a_k| \) for all positive integers \( k \) due to the properties of square roots and division. As \( k \) increases, its square root increases as well, hence the reciprocal decreases. This is an important property as it assures that later terms in the series will not inexplicably increase in magnitude, which could disrupt convergence.

Lastly, the concept of the limit of a sequence plays an essential role in determining convergence. \( \lim_{n \to \infty} a_n = L \) means that as \( n \) progresses towards infinity, the sequence \( a_n \) approaches the value \( L \) more closely. If \( L \) is a finite number, we say that the sequence has a limit \( L \) as \( n \) approaches infinity.

For the alternating series to converge, not only must the term sequence be decreasing, but the limit of the terms as \( n \) approaches infinity must be zero. This is sometimes referred to as the n-th term test for convergence. In our given series, the limit \( \lim_{k \to \infty} \frac{(-1)^{k}}{\sqrt{k}} = 0 \) as confirmed in our solution. In other words, the magnitude of the terms gets infinitesimally small as we move towards the end of the series. This second condition of the Alternating Series Test ensures that adding more terms contributes negligibly to the total sum, allowing for convergence.