91Ó°ÊÓ

Since we are interested in the sum for \(k \geq 10\), we will set up the integral as \(\int_{10}^{\infty} \frac{1}{x^{p}} dx\). This integral will tell us the convergence or divergence of the original series.

Evaluate the integral \(\int_{10}^{\infty} \frac{1}{x^{p}} dx\). To do so, we will first integrate it as a regular integral: \(\int \frac{1}{x^{p}} dx = \int x^{-p} dx = \frac{x^{-p + 1}}{-p + 1} + C = \frac{x^{1 - p}}{1 - p} + C\) Now, we need to determine the limits of integration: \(\lim_{b \rightarrow \infty} \int_{10}^{b} \frac{1}{x^{p}} dx = \lim_{b \rightarrow \infty} \left[ \frac{x^{1 - p}}{1 - p} \right]_{10}^{b}\)

Now we want to determine when this limit converges or diverges for different values of \(p\). This limit simplifies to: \(\lim_{b \rightarrow \infty} \left[ \frac{b^{1 - p}}{1 - p} - \frac{10^{1 - p}}{1 - p} \right]\) If \(p > 1\), then the term \(b^{1 - p}\) goes to zero as \(b \rightarrow \infty\) because \((1 - p) < 0\). In this case, the limit converges, and by the integral test, the series also converges. If \(p = 1\), then the term \(b^{1 - p}\) becomes \(b^{0} = 1\) and the integral diverges as \(b \rightarrow \infty\), since we have \(\int_{10}^{\infty} \frac{1}{x} dx\). By the integral test, the series diverges. If \(p < 1\), then the term \(b^{1 - p}\) increases without bound as \(b \rightarrow \infty\) because \((1 - p) > 0\). The limit and the integral diverge, and by the integral test, the series also diverges.

In conclusion, the series \(\sum_{k=10}^{\infty} \frac{1}{k^{p}}\) converges when \(p > 1\) and diverges when \(p \leq 1\).