91Ó°ÊÓ

In the given series, we can identify that the sequence \(a_k\) is given by \(a_k = k^{1/k}\). To check if the terms are decreasing, we need to check if \(a_k \ge a_{k+1}\) for all \(k\). We can check this condition by taking the derivative of the terms with respect to \(k\). Let's take the derivative of \(a_k\): $$\frac{d}{dk}k^{1/k} = \frac{d}{dk}(e^{\frac{\ln{k}}{k}})$$ We can apply the chain rule for derivatives here: $$\frac{d}{dk}k^{1/k} = e^{\frac{\ln{k}}{k}}\frac{d}{dk}\frac{\ln{k}}{k}$$ Now, we need to find the derivative of \(\frac{\ln{k}}{k}\) using the quotient rule: $$\frac{d}{dk}\frac{\ln{k}}{k} = \frac{k\frac{1}{k} - \ln{k}(1)}{k^2} = \frac{1-\ln{k}}{k^2}$$ The derivative of \(a_k\) is: $$\frac{d}{dk}k^{1/k} = e^{\frac{\ln{k}}{k}}\frac{1-\ln{k}}{k^2}$$ From the result, we can see that the derivative is negative when \(\ln{k}>1\) (i.e., \(k > e\)), which means that the terms are decreasing for \(k>e\). Since our series starts from \(k=1\), we can ignore the first few terms that do not satisfy this decreasing property, as it does not affect the convergence or divergence of the series. Thus, the first condition of the Alternating Series Test is met.

The next step is to check if the limit of the terms approaches zero, i.e., \(\lim_{k \to \infty} a_k = 0\). $$\lim_{k \to \infty} k^{1/k} = \lim_{x \to \infty} e^{\frac{\ln{k}}{k}}$$ We can use L'Hopital's rule on the exponent to find the limit: $$\lim_{k \to \infty} \frac{\ln{k}}{k} = \lim_{k \to \infty} \frac{\frac{1}{k}}{1} = \lim_{k \to \infty} \frac{1}{k} = 0$$ Since the limit of the exponent is zero, the limit of the terms is: $$\lim_{k \to \infty} k^{1/k} = e^0 = 1 \neq 0$$ The second condition of the Alternating Series Test is not met.

Since the second condition is not met, the given alternating series $$\sum_{k=1}^{\infty}(-1)^{k+1} k^{1 / k}$$ does not converge according to the Alternating Series Test.