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

Options

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

Answer

To solve the problem \((1/(2!) + 1/(4!) + 1/(6!) + \ldots) / (1 + 1/(3!) + 1/(5!) + \ldots)\), we will break it down step by step.

### Step 1: Identify the Series
The numerator is the series:
\[ S_1 = \sum_{n=1}^{\infty} \frac{1}{(2n)!} = \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \ldots \]
The denominator is the series:
\[ S_2 = 1 + \sum_{n=1}^{\infty} \frac{1}{(2n+1)!} = 1 + \frac{1}{3!} + \frac{1}{5!} + \ldots \]

### Step 2: Evaluate the Series
The series \(S_1\) can be recognized as half of the Taylor series expansion for \(e^x\) evaluated at \(x=1\):
\[ S_1 = \frac{1}{2} (e + e^{-1}) = \frac{1}{2} (e + \frac{1}{e}) = \frac{e + \frac{1}{e}}{2} \]

The series \(S_2\) can be evaluated as:
\[ S_2 = e + \sum_{n=1}^{\infty} \frac{1}{(2n)!} = e + \frac{1}{2}(e + \frac{1}{e}) - 1 = e - 1 + \frac{e + \frac{1}{e}}{2} \]
This simplifies to:
\[ S_2 = e - 1 + \frac{e + \frac{1}{e}}{2} \]

### Step 3: Simplifying the Expression
Now we need to simplify the expression:
\[ \frac{S_1}{S_2} = \frac{\frac{e + \frac{1}{e}}{2}}{e - 1 + \frac{e + \frac{1}{e}}{2}} \]

### Step 4: Multiply Numerator and Denominator by 2
To eliminate the fraction in the numerator, we multiply both the numerator and denominator by 2:
\[ \frac{e + \frac{1}{e}}{2(e - 1 + \frac{e + \frac{1}{e}}{2})} \]

### Step 5: Simplifying Further
The denominator simplifies to:
\[ 2(e - 1) + (e + \frac{1}{e}) = 2e - 2 + e + \frac{1}{e} = 3e - 2 + \frac{1}{e} \]
Thus, we have:
\[ \frac{e + \frac{1}{e}}{3e - 2 + \frac{1}{e}} \]

### Step 6: Final Simplification
Now we can factor out common terms:
\[ = \frac{e^2 + 1}{3e^2 - 2e + 1} \]

### Final Answer
The final answer is:
\[ \frac{e + 1}{e - 1} \]