91Ó°ÊÓ

The Ratio Test states that a series converges if $$\lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| < 1$$ and it diverges if $$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right|>1$$ where \(a_k\) represents the terms of the series. For the given series, we have: $$a_k = \frac{(k!)^2}{(2k)!}$$

We need to compute the ratio \(\frac{a_{k+1}}{a_k}\): $$\frac{a_{k+1}}{a_k} = \frac{\frac{((k+1)!)^2}{(2(k+1))!}}{\frac{(k!)^2}{(2k)!}}$$ Simplifying, we have: $$\frac{a_{k+1}}{a_k} = \frac{((k+1)!)^2(2k)!}{(k!)^2(2(k+1))!}$$

First, we can simplify \((2(k+1))!\) in the denominator: $$(2(k+1))! = (2k+2)! = (2k+2)(2k+1)...3\cdot2\cdot1$$ We can rewrite \((2k)!\) as: $$(2k)! = 2k(2k-1)...3\cdot2\cdot1$$ Now, we can simplify the ratio: $$\frac{a_{k+1}}{a_k} = \frac{((k+1)!)^2(2k)(2k-1)...3\cdot2\cdot1}{(k!)^2(2k+2)(2k+1)...3\cdot2\cdot1}$$ Divide both numerator and denominator by \((2k)(2k-1)...3\cdot2\cdot1\): $$\frac{a_{k+1}}{a_k} = \frac{((k+1)!)^2}{(k!)^2(2k+2)(2k+1)}$$

The ratio can now be rewritten as: $$\frac{a_{k+1}}{a_k} = \frac{((k+1)!)^2}{((k!)^2(2k+2)(2k+1)} = \frac{(k+1)^2((k!)^2)}{(2k+2)(2k+1)((k!)^2)}$$ Cancel out \((k!)^2\) from the numerator and denominator: $$\frac{a_{k+1}}{a_k} = \frac{(k+1)^2}{(2k+2)(2k+1)}$$ Now, we will find the limit of the ratio as \(k\) approaches infinity: $$\lim_{k \to \infty} \frac{(k+1)^2}{(2k+2)(2k+1)}$$ Divide both numerator and denominator by \(k^2\): $$\lim_{k \to \infty} \frac{\left(\frac{k+1}{k}\right)^2}{\left(\frac{2k+2}{k}\right)\left(\frac{2k+1}{k}\right)}$$ Now as \(k\) approaches infinity, each fraction inside the limit becomes: $$\lim_{k \to \infty} \frac{(1+1/k)^{2}}{(2+2/k)(2+1/k)}$$ As \(k\to \infty\), the limit becomes: $$\lim_{k \to \infty} \frac{(1+0)^2}{(2+0)(2+0)}=\frac{1}{4}$$

Since the limit of the ratio of consecutive terms, \(\lim_{k \to \infty} \frac{a_{k+1}}{a_k} = \frac{1}{4}<1\), by the Ratio Test, the given series converges.