91Ó°ÊÓ

First, let's check whether the series converges absolutely. We can do this by considering the absolute value of the given series: $$\sum_{k=1}^{\infty} \left|\frac{(-1)^{k} k^{2}}{\sqrt{k^{6}+1}}\right|$$ Which simplifies to: $$\sum_{k=1}^{\infty} \frac{k^{2}}{\sqrt{k^{6}+1}}$$ To find out whether this series converges, we can use the Comparison Test or Limit Comparison Test.

Let's use the Limit Comparison Test. Consider the limit of the ratio of the given series to a simpler one, say \(1/k^2\). $$\lim_{k\to\infty} \frac{\frac{k^{2}}{\sqrt{k^{6}+1}}}{\frac{1}{k^{2}}}$$ Simplifying the limit: $$\lim_{k\to\infty} \frac{k^4}{\sqrt{k^{6}+1}}$$ As \(k\) approaches infinity, we can neglect the constants in the denominator, which leads to: $$\lim_{k\to\infty} \frac{k^4}{k^3}$$ After simplifying, we get: $$\lim_{k\to\infty} k = \infty$$ Since the limit is infinity, there is no comparison that allows us to determine convergence. Thus, we can conclude that the series does not converge absolutely.

As we have determined that the given series does not converge absolutely, we would now check whether the series converges conditionally. We can do this using the Alternating Series Test. For the Alternating Series Test, if the sequence \(a_k = \frac{k^2}{\sqrt{k^6+1}}\) is positive, decreasing, and has a limit of 0 as \(k \to \infty\), then the series converges. First, let's observe that the sequence \(a_k\) is positive since both the numerator and denominator are positive. Next, we need to show that the sequence is decreasing. We can do this by checking whether the following inequality holds: \(a_{k+1} \le a_k\) Solving the inequality for the given sequence: $$\frac{(k+1)^2}{\sqrt{(k+1)^6+1}} \le \frac{k^2}{\sqrt{k^6+1}}$$ Since both the numerator and denominator are strictly positive, we can safely square both sides: $$\left(\frac{(k+1)^2}{\sqrt{(k+1)^6+1}}\right)^2 \le \left(\frac{k^2}{\sqrt{k^6+1}}\right)^2$$ Simplifying: $$(k+1)^4(k^6+1) \le k^4((k+1)^6 + 1)$$ Intuitively when \(k\to\infty\), the inequality holds, and hence the sequence is decreasing. Finally, let's check whether the sequence has a limit of 0: $$\lim_{k\to\infty} \frac{k^2}{\sqrt{k^6+1}} = 0$$ Since the sequence \(a_k\) is positive, decreasing, and has a limit of 0, the Alternating Series Test shows that the given series converges conditionally. #Conclusion# The given series \(\sum_{k=1}^{\infty} \frac{(-1)^{k} k^{2}}{\sqrt{k^{6}+1}}\) converges conditionally, as it does not converge absolutely, but it does converge when considering the alternating sign.