$$ \frac{2^{n}}{(2 n)!} $$. Slove by test rako

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UpStudy AI · Accepted Answer

Certainly! Let's analyze the series

$$
\sum_{n=0}^{\infty} \frac{2^{n}}{(2n)!}
$$

using the **Ratio Test** to determine its convergence.

## **Ratio Test Overview**

The **Ratio Test** states that for a series $$\sum a_n$$:

1. Compute the limit $$ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$.
2. 
   - If $$ L < 1 $$, the series **converges absolutely**.
   - If $$ L > 1 $$ or $$ L $$ is **infinite**, the series **diverges**.
   - If $$ L = 1 $$, the test is **inconclusive**.

## **Applying the Ratio Test**

Given the series:

$$
a_n = \frac{2^{n}}{(2n)!}
$$

Let's compute the limit $$ L $$:

$$
L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{2^{n+1}}{(2(n+1))!} \cdot \frac{(2n)!}{2^{n}} = \lim_{n \to \infty} \frac{2 \cdot 2^n}{(2n+2)(2n+1) \cdot 2n!} \cdot \frac{2n!}{2^n} 
$$

Simplifying the expression:

$$
L = \lim_{n \to \infty} \frac{2}{(2n+2)(2n+1)} = \lim_{n \to \infty} \frac{2}{4n^2 + 6n + 2}
$$

As $$ n \to \infty $$, the dominant term in the denominator is $$ 4n^2 $$, so:

$$
L = \lim_{n \to \infty} \frac{2}{4n^2} = \lim_{n \to \infty} \frac{1}{2n^2} = 0
$$

## **Conclusion**

Since $$ L = 0 < 1 $$, by the **Ratio Test**, the series

$$
\sum_{n=0}^{\infty} \frac{2^{n}}{(2n)!}
$$

**converges absolutely**.

## **Additional Insight**

It's interesting to note that the series

$$
\sum_{n=0}^{\infty} \frac{2^{n}}{(2n)!}
$$

is related to the hyperbolic cosine function, which is defined as:

$$
\cosh(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}
$$

For $$ x = \sqrt{2} $$, the series becomes:

$$
\cosh(\sqrt{2}) = \sum_{n=0}^{\infty} \frac{2^{n}}{(2n)!}
$$

Since the hyperbolic cosine function converges for all real numbers, this provides another confirmation of the convergence of the given series.
The series $$ \sum_{n=0}^{\infty} \frac{2^{n}}{(2n)!} $$ converges absolutely by the Ratio Test.

\( \frac{2^{n}}{(2 n)!} \). Slove by test rako

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