determine whether the following series converge. 24. ∑k = 2^∞ (-1)^k / (k ln²k)

Explanation:

Step1: Check alternating - series test conditions

The given series is $\sum_{k = 2}^{\infty}\frac{(- 1)^{k}}{k\ln^{2}k}$, which is an alternating series of the form $\sum_{k = 2}^{\infty}(-1)^{k}a_{k}$ where $a_{k}=\frac{1}{k\ln^{2}k}$. First, we need to show that $a_{k}>0$ for all $k\geq2$. Since $k > 1$ and $\ln k>0$ for $k > 1$, then $a_{k}=\frac{1}{k\ln^{2}k}>0$ for $k\geq2$.

Step2: Show $a_{k + 1}\leq a_{k}$

Let $f(x)=\frac{1}{x\ln^{2}x}$, and find its derivative using the quotient - rule. The quotient rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 1$, $u^\prime=0$, $v=x\ln^{2}x$, and $v^\prime=\ln^{2}x + 2x\frac{1}{x}\ln x=\ln^{2}x + 2\ln x$. So, $f^\prime(x)=\frac{0-( \ln^{2}x + 2\ln x)}{(x\ln^{2}x)^{2}}=-\frac{\ln x(\ln x + 2)}{x^{2}\ln^{4}x}=-\frac{\ln x + 2}{x^{2}\ln^{3}x}$. For $x\geq2$, $\ln x>0$, so $f^\prime(x)<0$. This means $f(x)$ is a decreasing function. So, $a_{k + 1}=f(k + 1)\leq a_{k}=f(k)$ for $k\geq2$.

Step3: Find the limit of $a_{k}$ as $k\to\infty$

We find $\lim_{k\to\infty}a_{k}=\lim_{k\to\infty}\frac{1}{k\ln^{2}k}$. Let $t=\ln k$, as $k\to\infty$, $t\to\infty$. Then the limit becomes $\lim_{t\to\infty}\frac{1}{e^{t}t^{2}}$. Using L'Hopital's rule twice. First, consider the limit $\lim_{t\to\infty}\frac{1}{e^{t}t^{2}}=\lim_{t\to\infty}\frac{0}{e^{t}(t^{2}+2t)} = 0$ (since the derivative of the numerator is 0 and the derivative of the denominator $e^{t}t^{2}$ using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ with $u = e^{t}$, $u^\prime=e^{t}$, $v = t^{2}$, $v^\prime = 2t$ is $e^{t}(t^{2}+2t)$). Since the three conditions of the alternating - series test are satisfied ($a_{k}>0$, $a_{k + 1}\leq a_{k}$, and $\lim_{k\to\infty}a_{k}=0$), the series $\sum_{k = 2}^{\infty}\frac{(-1)^{k}}{k\ln^{2}k}$ converges.

Answer:

The series $\sum_{k = 2}^{\infty}\frac{(-1)^{k}}{k\ln^{2}k}$ converges.