91Ó°ÊÓ

We need to show that the terms in the non-alternating series are non-increasing; that is, $$a_k \geq a_{k+1}$$. To do this, we will use the fact that the function $$f(x)=x^{1/x}$$ is a decreasing function for $$x \geq e$$: First, we will verify that $$f(x)=x^{1/x}$$ is decreasing for $$x \geq e$$ by computing its derivative, so $$f'(x)=\frac{d}{dx}(x^{1/x})$$. To compute the derivative, we need to use logarithmic differentiation: 1. Take the natural logarithm of both sides: $$\ln f(x)=\frac{1}{x} \ln x$$ 2. Compute the derivative of both sides with respect to $$x$$: $$\frac{f'(x)}{f(x)}=\frac{d}{dx} (\frac{1}{x}\ln x)$$ Now we will differentiate the expression and obtain: $$\frac{f'(x)}{f(x)}= -\frac{1}{x^2}\ln x + \frac{1}{x^2}$$ Then, $$f'(x)=(-\frac{1}{x^2}\ln x + \frac{1}{x^2})f(x)=(-\frac{1}{x^2}\ln x + \frac{1}{x^2})x^{1/x}$$ For $$x\geq e$$: $$-\frac{1}{x^2}\ln x + \frac{1}{x^2}<0$$ since $$\ln x \geq 1$$ when $$x\geq e$$. Now, for $$k=1$$, we have: $$k^{1/k}=1^{1/1}=1$$ and for $$k=2$$: $$k^{1/k}=2^{1/2}\approx 1.414$$ It is clear that since $$f(x)=x^{1/x}$$ is decreasing for $$x\geq e$$, the same can be said for $$k\geq 2$$, implying that $$a_k \geq a_{k+1}$$ for $$k\geq 2$$.

In order to apply the AST, we need to verify if $$\lim_{k\to\infty} a_k = \lim_{k\to\infty} k^{1/k}=0$$. To do this, we will utilize one of the properties of limits: if $$\lim_{k\to\infty} \frac{a_k}{b_k}=1$$ and $$\lim_{k\to\infty}b_k=0$$, then $$\lim_{k\to\infty}a_k=0$$. Let's rewrite the limit of $$a_k$$: $$\lim_{k\to\infty} k^{1/k}=\lim_{k\to\infty} \frac{e^{\ln(k^{1/k})}}{1}=\lim_{k\to\infty}\frac{e^{\frac{1}{k}\ln k}}{1}$$ Define $$b_k=e^{\frac{1}{k}\ln k}$$, and now we can rewrite the limit as: $$\lim_{k\to\infty} \frac{a_k}{b_k}=\lim_{k\to\infty} \frac{e^{-\frac{1}{k}\ln k +\frac{1}{k}\ln k}}{1/e^{\frac{1}{k}\ln k}}=1$$ The limit $$\lim_{k\to\infty}b_k=\lim_{k\to\infty}e^{\frac{1}{k}\ln k}=1$$, which is not 0, hence we can't claim $$\lim_{k\to\infty}a_k=0$$. Since the second condition of the AST is not met, we cannot confirm the convergence of the series. So, the given series does not converge.