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To solve the given expression, we need to identify the series in both the numerator and the denominator and evaluate them.

The series are as follows:

Numerator:
\[ \sum_{n=1}^{\infty} \frac{1}{(2n)!} = \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \cdots \]

Denominator:
\[ \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} = \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \cdots \]

### Analyzing the Series

These series are parts of the Taylor expansion of the exponential function \(e^x\), which is given by:

\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]

We can decompose this into even and odd parts. 

### Even Part
The even part of the exponential series is:
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \]

For \( x = 1 \):

The even part:
\[ \sum_{n=0}^{\infty} \frac{1^{2n}}{(2n)!} = \sum_{n=0}^{\infty} \frac{1}{(2n)!} \]

### Odd Part
The odd part:
\[ \sum_{n=0}^{\infty} \frac{1^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} \]

Thus, the numerator of the given series is the even part of \( e^x \) evaluated at \( x=1 \), and the denominator is the odd part of \( e^x \) evaluated at \( x=1 \).

### Evaluating the Series
The numerator:
\[ \sum_{n=0}^{\infty} \frac{1}{(2n)!} \]

The denominator:
\[ \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} \]

So, the numerator is \( \cosh(1) \), and the denominator is \( \sinh(1) \), where \(\cosh\) and \(\sinh\) are the hyperbolic cosine and sine functions, respectively.

We know that:
\[ e = \cosh(1) + \sinh(1) \]

Therefore:
\[ \cosh(1) = \frac{e + e^{-1}}{2} \]
\[ \sinh(1) = \frac{e - e^{-1}}{2} \]

Now, the ratio of the two sums is:
\[ \frac{\sum_{n=0}^{\infty} \frac{1}{(2n)!}}{\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}} = \frac{\cosh(1)}{\sinh(1)} = \frac{\frac{e + e^{-1}}{2}}{\frac{e - e^{-1}}{2}} = \frac{e + e^{-1}}{e - e^{-1}} \]

Simplifying this, we get:
\[ \frac{e + e^{-1}}{e - e^{-1}} = \frac{e(1 + e^{-2})}{e(1 - e^{-2})} = \frac{1 + e^{-2}}{1 - e^{-2}} = \frac{e^2 + 1}{e^2 - 1} = \frac{e^2 + 1}{e^2 - 1} \]

By multiplying the numerator and denominator by \( e^2 \):

\[ \frac{e^2 + 1}{e^2 - 1} \]

Therefore, the correct option is:

(B) \(\frac{e + 1}{e - 1}\)

Would you like more details on any part of this solution or have any questions?

Here are some questions you might consider asking next:
1. How do you derive the Taylor series expansions for exponential functions?
2. What are the properties of hyperbolic functions?
3. Can you explain more about the relationship between hyperbolic and exponential functions?
4. How can series be transformed to find easier solutions?
5. What are some other common series involving factorials?

**Tip:** Understanding the connection between series and their related functions can help simplify complex problems significantly.

Learn how to solve a series problem using Taylor series expansion and hyperbolic functions. This solution covers the relationship between series and exponential functions, specifically using hyperbolic cosine and sine functions. Suitable for advanced mathematics students.

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