The problem asks us to determine if the series
$\sum_{n=1}^\infty \frac{(-1)^n}{(2n+1)!}$
converges using the D'Alembert Ratio Test (Признак Даламбера).

Step 1: Apply the Ratio Test

The D'Alembert Ratio Test states that for a series $\sum a_n$, where $a_n$ are terms of the series:

$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$

• If $L < 1$, the series converges absolutely.
• If $L > 1$ or $L = \infty$, the series diverges.
• If $L = 1$, the test is inconclusive.

Here, the terms $a_n$ of the series are:
$a_n = \frac{(-1)^n}{(2n+1)!}.$

Step 2: Find $\frac{a_{n+1}}{a_n}$:

We calculate the ratio:

$\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{(-1)^{n+1}}{(2(n+1)+1)!}}{\frac{(-1)^n}{(2n+1)!}} \right|$

Simplifying the expression:

$\left| \frac{a_{n+1}}{a_n} \right| = \frac{(2n+1)!}{(2(n+1)+1)!}.$

Simplify the factorial terms:

• The numerator is $(2n+1)!$.
• The denominator is $(2n+3)!$, which expands as:
  $(2n+3)! = (2n+3)(2n+2)(2n+1)!.$

Thus,

$\left| \frac{a_{n+1}}{a_n} \right| = \frac{(2n+1)!}{(2n+3)(2n+2)(2n+1)!}.$

Canceling $(2n+1)!$ in numerator and denominator:

$\left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{(2n+3)(2n+2)}.$

Step 3: Take the limit as $n \to \infty$:

As $n \to \infty$, the product $(2n+3)(2n+2) \to \infty$, so:

$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0.$

Step 4: Conclusion

Since $L = 0 < 1$, the series converges absolutely.

Final Answer:

The number to enter is 1 (converges absolutely).

Would you like me to explain any specific part further or clarify the steps?

Related Questions

1. What is the D'Alembert Ratio Test, and how is it applied?
2. What does absolute convergence mean?
3. How does the factorial $n!$ behave as $n \to \infty$?
4. Why does $\frac{1}{(2n+3)(2n+2)} \to 0$ as $n \to \infty$?
5. What is the difference between absolute and conditional convergence?

Tip: When handling factorials, always look for cancellation to simplify the expressions.