91Ó°ÊÓ

Since \((1+\frac{1}{k})>1\) for all \(k\), we have $$a_k=\left(1+\frac{1}{k}\right)^{k}>1^k=1$$ Thus, \(a_k > 0\) for all \(k\).

To determine whether the sequence is decreasing, we can check if: $$\frac{a_{k+1}}{a_k} \le 1$$ for all \(k\). Now, $$\frac{a_{k+1}}{a_k} = \frac{\left(1+\frac{1}{k+1}\right)^{k+1}}{\left(1+\frac{1}{k}\right)^{k}} = \left(\frac{1+\frac{1}{k+1}}{1+\frac{1}{k}}\right)^{k+1}.$$ Let \(y_k = \frac{k}{k+1}\). Then \(1-y_k = \frac{1}{k+1}\) and $$\frac{a_{k+1}}{a_k} = \left(\frac{1+\frac{1}{k+1}}{1+\frac{1}{k}}\right)^{k+1} = \frac{(1-y_k)^{-1}}{(1-y_{k+1})^{-1}}^{k+1} = \left(\frac{1-y_{k+1}}{1-y_k}\right)^{k+1} < e^{\frac{y_{k+1}-y_k}{1-y_{k+1}}*(k+1)}.$$ Now, by applying Bernoulli's Inequality, we have \((1+x)^n \ge 1+nx\) for all \(x > -1\) and \(n \in \mathbb{N}\). Applying this result to our expression, we obtain $$a_{k+1} \le e^{\frac{y_{k+1}-y_k}{1-y_{k+1}}*(k+1)} \le 1+\frac{(y_{k+1}-y_k)(k+1)}{1-y_{k+1}}.$$ Now, \(y_{k+1}-y_k = \frac{-1}{k+1}*\frac{-1}{k}\), and we know that \(y_{k+1} < y_k\), therefore $$a_{k+1} \le 1+\frac{(y_{k+1}-y_k)(k+1)}{1-y_{k+1}} \le 1+\frac{(y_{k+1}-y_k)(k+1)}{1-y_k} \le 1$$. It follows that \(\frac{a_{k+1}}{a_k} \le 1\) for all \(k\), and therefore \(a_k\) is a decreasing sequence.

We know that $$a_k=\left(1+\frac{1}{k}\right)^{k}.$$ We can rewrite this as: $$a_k=e^{k\ln(1+\frac{1}{k})}$$ Applying L'Hospital's Rule to the limit, $$\lim\limits_{k\to\infty} k\ln(1+\frac{1}{k}) = \lim\limits_{k\to\infty} \frac{\ln(1+\frac{1}{k})}{\frac{1}{k}}.$$ Now applying L'Hospital's Rule again, $$\lim\limits_{k\to\infty} \frac{\frac{-1}{k^2(1+\frac{1}{k})}}{\frac{-1}{k^2}} = \lim\limits_{k\to\infty} 1+\frac{1}{k} = 1.$$ So, $$\lim\limits_{k\to\infty} a_k = e^1 = e \neq 0.$$ Since conditions 1 and 2 are satisfied, but condition 3 is not, we conclude that the series \(\sum_{k=1}^{\infty}(-1)^{k}\left(1+\frac{1}{k}\right)^{k}\) does not converge according to the Alternating Series Test.