91Ó°ÊÓ

We are given a sequence with terms defined as: $$ a_{2k-1} = a^{k-1}b^{k-1} \quad \text{and} \quad a_{2k} = a^kb^{k-1} $$ for \(k \in \mathbb{N}\). These terms alternate in the series \(\sum_{k \geq 1}a_k\).

Recall that the Ratio Test states that, given a series \(\sum_{n = 1}^{\infty}c_n\): - if \(\lim_{n\to\infty} |\frac{c_{n+1}}{c_n}| < 1\), the series converges, - if \(\lim_{n\to\infty} |\frac{c_{n+1}}{c_n}| > 1\), the series diverges, and - if \(\lim_{n\to\infty} |\frac{c_{n+1}}{c_n}| = 1\), the test is inconclusive. Here, we will analyze the limit for the given series \(\sum_{k\geq 1}a_k\). We must analyze the ratio of consecutive terms for even \(n\) (corresponding to odd \(k\)) and odd \(n\) (corresponding to even \(k\)) separately. 1. When \(n = 2k - 1\): $$ \lim_{k\to\infty}\left|\frac{a_{2k}}{a_{2k-1}}\right| = \lim_{k\to\infty}\left|\frac{a^k b^{k-1}}{a^{k-1} b^{k-1}}\right| = \lim_{k\to\infty} |a| $$ Since \(0 < a < b\), if \(a < 1\) then the limit is less than 1, which means that the series converges. If \(a \geq 1\), the limit is greater than or equal to 1, which means that the series diverges. 2. When \(n = 2k\): $$ \lim_{k\to\infty}\left|\frac{a_{2k+1}}{a_{2k}}\right| = \lim_{k\to\infty}\left|\frac{a^{k} b^{k}}{a^{k} b^{k-1}}\right| = \lim_{k\to\infty} |b| $$ Since \(0 < a < b\), if \(b < 1\) then the limit is less than 1, which means that the series converges. If \(b \geq 1\), the limit is greater than or equal to 1, which means that the series diverges. Putting these two cases together, for the series to converge, we must have \(a<1\) and \(b<1\).

Recall that the Root Test states that, given a series \(\sum_{n = 1}^{\infty}c_n\): - if \(\lim_{n\to\infty} |{c_n}^{\frac{1}{n}}| < 1\), the series converges, - if \(\lim_{n\to\infty} |{c_n}^{\frac{1}{n}}| > 1\), the series diverges, and - if \(\lim_{n\to\infty} |{c_n}^{\frac{1}{n}}| = 1\), the test is inconclusive. Again, we must analyze the given series \(\sum_{k\geq 1}a_k\) for even and odd \(n\) separately. 1. When \(n = 2k - 1\): $$ \lim_{k\to\infty}\left| a_{2k-1}^{\frac{1}{2k-1}}\right| = \lim_{k\to\infty} | a^{(k-1)/(2k-1)}b^{(k-1)/(2k-1)} | $$ Since both \(a\) and \(b\) are raised to fractions that converge to \(0\) (as \(k\to\infty\)), the limit converges to 1. 2. When \(n = 2k\): $$ \lim_{k\to\infty}\left| a_{2k}^{\frac{1}{2k}}\right| = \lim_{k\to\infty} | a^{k/(2k)}b^{(k-1)/(2k)} | = \lim_{k\to\infty} | a^{1/2}b^{(k-1)/(2k)} | $$ Again, the exponent for \(b\) converges to 0 as \(k\to\infty\). So, the limit reduces to: $$ \lim_{k\to\infty} | a^{1/2} | = |a|^{1/2} $$ For the series to converge, we require \(|a|^{1/2} < 1\), which means that the product \(ab<1\). Conversely, if \(ab > 1\), the series diverges. Hence, the series converges if \(ab < 1\).