91Ó°ÊÓ

In the given series, we can see that there is a term \((-1)^{k+1}\), indicating that it is an alternating series. Let's first rewrite the series with the terms a_n as follows: $$\sum_{k=1}^{\infty}(-1)^{k+1} \frac{k^{2}}{k^{3}+1} = \sum_{k=1}^{\infty}(-1)^{k+1}a_k,$$ where \(a_k = \frac{k^{2}}{k^{3}+1}\).

To determine whether the alternating series converges, we will apply the Alternating Series Test (AST): 1. The terms \(a_k\) are non-increasing (\(a_{k+1} \leq a_k\) for each \(k\)). 2. \(\lim_{k \to \infty}a_k = 0\). If both of these conditions are met, the alternating series converges.

First, we need to show that the terms \(a_k\) are non-increasing. We can do this by checking the derivative of \(a_k\): \(a_k = \frac{k^{2}}{k^{3}+1}\) Let's find the derivative of \(a_k\) with respect to \(k\). $$\frac{d}{dk}\left(\frac{k^{2}}{k^{3}+1}\right)$$ Applying the quotient rule, we have: $$\frac{d}{dk}\left(\frac{k^{2}}{k^{3}+1}\right) = \frac{2k(k^{3}+1) - 3k^{3}k^{2}}{(k^{3}+1)^{2}} = \frac{2k^4 -3k^5 + 2k}{(k^3+1)^2}$$ Notice that the numerator of the above expression is a polynomial with a leading term of \(-k^5\). As \(k\) becomes larger, the term \(-k^5\) dominates the numerator, so the overall expression becomes negative. Thus, the derivative of \(a_k\) is negative, implying that the terms of \(a_k\) are non-increasing for all positive values of \(k\).

Now, we will check if \(\lim_{k \to \infty}a_k = 0\): $$\lim_{k \to \infty} \frac{k^{2}}{k^{3}+1}$$ Dividing all terms in the numerator and denominator by \(k^3\), we obtain: $$\lim_{k \to \infty} \frac{\frac{k^2}{k^3}}{\frac{k^3}{k^3} + \frac{1}{k^3}} = \lim_{k \to \infty} \frac{\frac{1}{k}}{1 + \frac{1}{k^3}}$$ As \(k\) approaches \(\infty\), the terms with \(k\) in the denominator approach 0, which gives: $$\lim_{k \to \infty} \frac{0}{1+0} = 0$$

Since both the conditions of the Alternating Series Test (AST) are met (the terms \(a_k\) are non-increasing and \(\lim_{k \to \infty}a_k = 0\)), we can conclude that the alternating series converges. Therefore, the given series $$\sum_{k=1}^{\infty}(-1)^{k+1} \frac{k^{2}}{k^{3}+1}$$ converges.