Question

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n !}{1 \cdot 3 \cdot 5 \cdot \cdot \cdot(2 n-1)} $$

Short answer

The given series converges.

Step-by-step solution

Apply the Ratio Test

The Ratio Test requires to find the limit as \( n \) approaches infinity of the absolute value of the ratio of the \( n+1 \) term to the \( n \) term of the series.\n Thus, apply the ratio test which states:\[\lim_{{n \to \infty}} \frac{a_{n+1}}{a_n} = L\]The terms of the series \( a_n \) and \( a_{n+1} \) are:\[a_n= \frac{(-1)^{n+1} n !}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2 n-1)}\]\[a_{n+1}= \frac{(-1)^{n+2} (n+1) !}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n+1)}\]

Calculate the limit

Divide \( a_{n+1} \) by \( a_n \) and apply limit as \( n \) approaches infinity.\[L= \lim_{n \to \infty} \left| \frac{(-1)^{n+2} (n+1) ! / 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2 n+1)}{(-1)^{n+1} n ! / 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2 n-1)}\right|\]Simplify\[L= \lim_{n \to \infty} \left| \frac{(-1)(n+1)(2n-1)!}{(2n+1)!}\right|\]\[L= \lim_{n \to \infty} \left| \frac{-1}{2n+1}\right|\]

Interpret the result

Since the absolute value of \( L \) is zero (which is less than 1), the given series converges by the Ratio Test. Hence, it results in an alternating series, each of whose terms is smaller than the one before.