Questions: ∑(k=2 to ∞) (k-1) / √(k^3+k+1)

Solution Steps

To determine the convergence of the series \(\sum_{k=2}^{\infty} \frac{k-1}{\sqrt{k^{3}+k+1}}\), we can use the Limit Comparison Test. We compare the given series with a simpler series that we know the convergence properties of, such as \(\sum_{k=2}^{\infty} \frac{1}{k^{3/2}}\), which is a p-series with \(p = 3/2 > 1\) and is known to converge. We then compute the limit of the ratio of the terms of the two series as \(k\) approaches infinity.

We are given the series

\[ \sum_{k=2}^{\infty} \frac{k-1}{\sqrt{k^{3}+k+1}}. \]

To analyze its convergence, we will use the Limit Comparison Test with the series

\[ \sum_{k=2}^{\infty} \frac{1}{k^{3/2}}. \]

We define

\[ a_k = \frac{k-1}{\sqrt{k^{3}+k+1}} \quad \text{and} \quad b_k = \frac{1}{k^{3/2}}. \]

Next, we compute the limit of the ratio of \(a_k\) to \(b_k\) as \(k\) approaches infinity:

\[ \lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{\frac{k-1}{\sqrt{k^{3}+k+1}}}{\frac{1}{k^{3/2}}} = \lim_{k \to \infty} \frac{(k-1) k^{3/2}}{\sqrt{k^{3}+k+1}}. \]

After simplification, we find that this limit approaches infinity:

\[ \lim_{k \to \infty} \frac{a_k}{b_k} = \infty. \]

Since

\[ \lim_{k \to \infty} \frac{a_k}{b_k} = \infty \]

and the series

\(\sum_{k=2}^{\infty} \frac{1}{k^{3/2}}\) converges (as it is a p-series with \(p = \frac{3}{2} > 1\)), we conclude by the Limit Comparison Test that the original series

\[ \sum_{k=2}^{\infty} \frac{k-1}{\sqrt{k^{3}+k+1}} \]

also converges.

Final Answer