91Ó°ÊÓ

Since the factorials \(k!\) are always positive and \(k^k\) is also always positive, we can conclude that \(a_k = \frac{k !}{k^{k}}\) is positive for all \(k \in \mathbb{N}\). Now let's check if the terms are decreasing: $$\frac{a_k}{a_{k+1}}=\frac{\frac{k!}{k^k}}{\frac{(k+1)!}{(k+1)^{(k+1)}}}=\frac{k! (k+1)^{(k+1)}}{(k+1)!k^k}$$ After simplifying, $$\frac{a_k}{a_{k+1}}=\frac{(k+1)^k}{k^k} \Rightarrow \frac{(k+1)^k}{k^{k+1}}$$ Now, consider the ratio of consecutive terms in the series: $$\frac{(k+1)^k}{k^{k+1}} > 1 \Leftrightarrow \frac{(k+1)^k}{k^k} > 1 \Leftrightarrow \left(\frac{k+1}{k}\right)^k > \left(\frac{k}{k+1}\right)$$ Since \(\left(\frac{k+1}{k}\right)^k\) is an increasing function of \(k\), and as \(k \to \infty\), \(\left(\frac{k+1}{k}\right)^k \to e\), we have that for large \(k\): $$\left(\frac{k+1}{k}\right)^k > \left(\frac{k}{k+1}\right)$$ Thus, the terms \(a_k\) are decreasing.

Now, let's compute the limit of the sequence: $$\lim\limits_{k \to \infty} a_k = \lim\limits_{k \to \infty} \frac{k !}{k^{k}}$$ We can use Stirling's approximation for large \(k\) values: $$k! \sim \sqrt{2\pi k}\left(\frac{k}{e}\right)^k$$ By substituting the Stirling's approximation into the limit: $$\lim\limits_{k\to\infty}\frac{\sqrt{2\pi k}\left(\frac{k}{e}\right)^k}{k^{k}}=\lim\limits_{k\to\infty}\sqrt{2\pi k}\left(\frac{k}{e}\right)^{-k}$$ As \(k \to \infty\), we have: $$\lim\limits_{k\to\infty}\sqrt{2\pi k}\left(\frac{k}{e}\right)^{-k} = 0$$ Thus, we have verified that \(\lim\limits_{k \to \infty} a_k = 0\).

Since the terms of the sequence \(a_k = \frac{k !}{k^{k}}\) are positive, decreasing, and \(\lim\limits_{k \to \infty} a_k = 0\), by the Alternating Series Test, the series: $$\sum_{k=1}^{\infty}(-1)^{k+1} \frac{k !}{k^{k}}$$ converges.