Questions: Consider the series sum from k=1 to infinity of (3+3k) / ((2+k^2)^2). a) Enter a p-series that can be used to determine the convergence or divergence of the given series by the Limit Comparison Test. To enter the series sum from k=1 to infinity of 1 / k^p type sum( k, 1, infinity, 1 / k^p ).

Solution Steps

To determine the convergence or divergence of the given series using the Limit Comparison Test, we need to find a $p$-series that behaves similarly to the given series for large $k$. The dominant term in the denominator of the given series is $k^4$, so we compare it to the series $\sum_{k=1}^{\infty} \frac{1}{k^4}$, which is a convergent $p$-series with $p = 4$.

To determine the convergence of the series \(\sum_{k=1}^{\infty} \frac{3+3k}{(2+k^2)^2}\), we first identify the dominant term in the denominator for large \(k\). The term \(k^2\) in the denominator becomes significant, making the denominator approximately \(k^4\).

We choose the \(p\)-series \(\sum_{k=1}^{\infty} \frac{1}{k^4}\) for comparison. This is a convergent series because \(p = 4 > 1\).

We apply the Limit Comparison Test by evaluating the limit:

\[ \lim_{k \to \infty} \frac{\frac{3+3k}{(2+k^2)^2}}{\frac{1}{k^4}} = \lim_{k \to \infty} \frac{(3+3k)k^4}{(2+k^2)^2} \]

The limit evaluates to \(\infty\), indicating that the given series behaves similarly to the comparison series for large \(k\).

Since the limit is \(\infty\) and the comparison series \(\sum_{k=1}^{\infty} \frac{1}{k^4}\) is convergent, the given series \(\sum_{k=1}^{\infty} \frac{3+3k}{(2+k^2)^2}\) is also convergent by the Limit Comparison Test.

Final Answer