91Ó°ÊÓ

Let's check if the function \(f(k)=\frac{k}{\left(k^{2}+1\right)^{3}}\) is continuous, positive, and decreasing for \(k\geq1\). 1. Continuity: The function is a rational function, and since the denominator is never zero, it is continuous for all \(k\geq1\). 2. Positivity: As \(k\geq1\) and the denominator is always positive, the function is positive for all \(k\geq1\). 3. Decreasing: To check if the function is decreasing, we can evaluate its derivative and check if it is negative for \(k\geq1\). Calculate the derivative of \(f(k)\) using the Quotient Rule: $$f'(k) = \frac{(k^2+1)^3(1) - k(6(k^2+1)^2(2k))}{(k^2+1)^6}$$ Notice that the denominator is always positive. We can now simplify the numerator: $$f'(k)=\frac{k^6+3k^4+3k^2+1-12k^5-12k^3}{(k^2+1)^6}$$ Let's check the sign of the numerator for \(k\geq1\). The terms \(k^6\) and \(12k^5\) are the most significant. As \(k\geq1\), we have \(k^6 \leq 12k^5\), which indicates that the numerator is less than or equal to zero. Therefore, the derivative \(f'(k)\leq0\) for \(k\geq1\). Since the function \(f(k)\) is continuous, positive, and decreasing for \(k\geq1\), we can apply the Integral Test.

Now, we need to evaluate the integral to determine the convergence or divergence of the series: $$\int_1^{\infty} \frac{k}{\left(k^{2}+1\right)^{3}} dk$$ To solve the integral, we can use substitution. Let \(u = k^2 + 1\), then \(du = 2k\,dk\). And also, replace the limits of integration: When \(k=1\), \(u=(1)^2+1=2\). When \(k\to\infty\), \(u\to\infty\). Substitute \(u\) and change the limits of integration: $$\frac{1}{2} \int_2^{\infty} \frac{1}{u^3} du$$ Now integrate with respect to \(u\): $$\frac{1}{2}\left[-\frac{1}{2u^2}\right]_2^{\infty}$$ $$\frac{1}{2}\left(-\frac{1}{2\infty^2}+\frac{1}{2(2)^2}\right) = \frac{1}{8}$$ Since the integral converges, the given series converges by the Integral Test.

By applying the Integral Test on the series \(\sum_{k=1}^{\infty} \frac{k}{\left(k^{2}+1\right)^{3}}\), we can conclude that the series is convergent.