91Ó°ÊÓ

We need to calculate the Ratio of consecutive terms of the series, i.e., \(\lim_{k\rightarrow\infty} \frac{a_{k+1}}{a_{k}}\) and apply the Ratio Test. Let's find the ratio of consecutive terms for alternating sequences: If \(k\) is odd, we have: \[\frac{a_{k+1}}{a_{k}}=\frac{a^{k+1} b^{k-1}}{a^{k-1}b^{k-1}}=a^2\] If \(k\) is even, we have: \[\frac{a_{k+1}}{a_{k}}=\frac{a^{k-1} b^{k}}{a^{k}b^{k-1}}=b\] Now applying the Ratio Test, we get: - If \(a \geq 1\), the \(\lim_{k\rightarrow\infty} \frac{a_{k+1}}{a_{k}}\geq 1\). Hence, the series is divergent. - If \(b < 1\), the \(\lim_{k\rightarrow\infty} \frac{a_{k+1}}{a_{k}} < 1\). Hence, the series is convergent. Thus, using the Ratio Test, we have shown that the series \(\sum_{k=1}^{\infty} a_{k}\) is convergent if \(b<1\), and it is divergent if \(a \geq 1\).

We need to calculate the root of the terms of the series, i.e., \(\lim_{k\rightarrow\infty}(a_k)^{\frac{1}{k}}\) and apply the Root Test. Let's find the root of the terms for alternating sequences: If \(k\) is odd, we have: \[(a_k)^{\frac{1}{k}}= (a^{k-1}b^{k-1})^{\frac{1}{k}}=((ab)^k)^{\frac{1}{2k}}=(ab)^\frac{1}{2}\] If \(k\) is even, we have: \[(a_k)^{\frac{1}{k}}= (a^kb^{k-1})^{\frac{1}{k}}= a(ab)^{\frac{k-1}{k}}\rightarrow a, \quad\text{as}\quad k\rightarrow \infty\] Now applying the Root Test, we get: - If \(ab < 1\), the \(\lim_{k\rightarrow\infty} (a_k)^{\frac{1}{k}} < 1\). Hence, the series is convergent. - If \(ab > 1\), the \(\lim_{k\rightarrow\infty} (a_k)^{\frac{1}{k}} > 1\). Hence, the series is divergent. Thus, using the Root Test, we have shown that the series \(\sum_{k=1}^{\infty} a_{k}\) is convergent if \(a b<1\), and it is divergent if \(a b>1\).