91Ó°ÊÓ

We can compare our given series with a p-series, which has the general form: $$\sum_{k=1}^{\infty} \frac{1}{k^p}.$$ In our case, the best p-series to compare would be with p = 3/2, as we have the term $$k^{3 / 2}$$ in our given series. So, we will compare our series with $$\sum_{k=1}^{\infty} \frac{1}{k^{3 / 2}}.$$

Let $$a_k = \frac{1}{k^{3 / 2}+1}$$ and $$b_k = \frac{1}{k^{3 / 2}}.$$ We will now use the Limit Comparison Test, which requires finding the limit $$\lim_{k \to \infty} \frac{a_k}{b_k}.$$ If the limit exists and is non-zero, then both series will have the same behavior (either both converge or both diverge). So let's compute this limit: $$\lim_{k \to \infty} \frac{\frac{1}{k^{3 / 2}+1}}{\frac{1}{k^{3 / 2}}} = \lim_{k \to \infty} \frac{k^{3 / 2}}{k^{3 / 2}+1}$$ To find this limit, we will divide the numerator and denominator by $$k^{3 / 2}.$$ Doing so, we get: $$\lim_{k \to \infty} \frac{k^{3 / 2}}{k^{3 / 2}+1} \cdot \frac{\frac{1}{k^{3 / 2}}}{\frac{1}{k^{3 / 2}}} = \lim_{k \to \infty} \frac{1}{1+\frac{1}{k^{3 / 2}}}.$$ As $$k$$ goes to infinity, the term $$\frac{1}{k^{3 / 2}}$$ goes to 0, so the limit becomes: $$\lim_{k \to \infty} \frac{1}{1+\frac{1}{k^{3 / 2}}} = \frac{1}{1+0} = 1 .$$ Since the limit is 1, a non-zero constant, both series have the same behavior. We know that the p-series $$\sum_{k=1}^{\infty} \frac{1}{k^{3 / 2}}$$ converges, because p > 1. Therefore, the given series $$\sum_{k=1}^{\infty} \frac{1}{k^{3 / 2}+1}$$ also converges. By using the Limit Comparison Test with a p-series of p = 3/2, we have shown that the given series converges.