`((1 )/(2!) + (1)/(4!) + (1)/(6!) + ......)/((1)/(1!) + (1)/(3!) + (1)/(5!) + ......)=`

Options

A. `(1 - e)/(1 + e)`
B. `(e - 1)/(e + 1)`
C. `(e + 1)/(1 - e)`
D. `(e + 1)/(e -1)`

Answer

To solve the given problem, we need to evaluate the expression:

\[ \frac{\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \ldots}{\frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \ldots} \]

### Step 1: Identify the series in the numerator and denominator

The numerator is the sum of the series for even factorials, and the denominator is the sum of the series for odd factorials.

### Step 2: Use the series expansion for \( e^x \) and \( e^{-x} \)

Recall the Taylor series expansions:
- \( e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \)
- \( e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots \)

### Step 3: Add \( e^x \) and \( e^{-x} \)

Adding these two series gives:
\[ e^x + e^{-x} = 2 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \]
This series contains only even factorials.

### Step 4: Substitute \( x = 1 \)

By substituting \( x = 1 \):
\[ e + e^{-1} = 2 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \ldots \]
Thus,
\[ \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \ldots = \frac{e + e^{-1} - 2}{2} \]

### Step 5: Subtract \( e^x \) and \( e^{-x} \)

Now, consider the difference:
\[ e^x - e^{-x} = \frac{x}{1!} + \frac{x^3}{3!} + \ldots \]
This series contains only odd factorials.

### Step 6: Substitute \( x = 1 \)

By substituting \( x = 1 \):
\[ e - e^{-1} = \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \ldots \]
Thus,
\[ \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \ldots = e - e^{-1} \]

### Step 7: Substitute the results back into the original expression

Now we can substitute these results into the original expression:
\[ \frac{\frac{e + e^{-1} - 2}{2}}{e - e^{-1}} \]

### Step 8: Simplify the expression

This simplifies to:
\[ \frac{e + e^{-1} - 2}{2(e - e^{-1})} \]

### Step 9: Multiply numerator and denominator by \( e \)

To simplify further, multiply the numerator and denominator by \( e \):
\[ \frac{(e^2 + 1 - 2e)}{2(e^2 - 1)} \]

### Step 10: Factor the numerator and denominator

The numerator can be factored as:
\[ (e - 1)^2 \]
And the denominator can be factored using the difference of squares:
\[ (e - 1)(e + 1) \]

### Step 11: Cancel the common terms

After canceling \( (e - 1) \):
\[ \frac{e - 1}{2(e + 1)} \]

### Final Result

Thus, the final result is:
\[ \frac{e - 1}{e + 1} \]

### Conclusion

The correct answer is option B:
\[ \frac{e - 1}{e + 1} \]

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