`sum_(n=1) ^(oo) (1)/((2n-1)!)= `

Options

A. `(e+e^(-1))/2`
B. `(e-e^(-1))/2`
C. `e/3`
D. `1/(3e)`

Answer

To solve the problem \( \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} \), we can follow these steps:

### Step 1: Understand the Series
The series we need to evaluate is:
\[ S = \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} \]
This series sums the reciprocals of the factorials of all odd integers.

### Step 2: Relate to the Exponential Function
Recall the Taylor series expansion for the exponential function \( e^x \):
\[ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \]
We can separate this into even and odd terms:
\[ e^x = \sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!} + \sum_{k=0}^{\infty} \frac{x^{2k+1}}{(2k+1)!} \]
The first sum represents the even factorials and the second sum represents the odd factorials.

### Step 3: Evaluate the Series for Specific Values
Let’s evaluate \( e^x \) at \( x = 1 \) and \( x = -1 \):
1. For \( x = 1 \):
\[ e^1 = \sum_{k=0}^{\infty} \frac{1^k}{k!} = \sum_{k=0}^{\infty} \frac{1}{k!} \]
2. For \( x = -1 \):
\[ e^{-1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \]

### Step 4: Combine the Results
Now, we can combine these two results:
\[ e^1 + e^{-1} = \sum_{k=0}^{\infty} \frac{1}{(2k)!} + \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} \]
The left-hand side gives us:
\[ e + \frac{1}{e} \]
The right-hand side separates into even and odd factorials:
\[ \sum_{k=0}^{\infty} \frac{1}{(2k)!} + \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} \]

### Step 5: Isolate the Odd Factorials
To isolate the odd factorials, we can subtract the even series from the total:
\[ \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} = e + \frac{1}{e} - \sum_{k=0}^{\infty} \frac{1}{(2k)!} \]

### Step 6: Find the Even Factorial Series
The even factorial series can be expressed as:
\[ \sum_{k=0}^{\infty} \frac{1}{(2k)!} = \cosh(1) \]
Thus, we can write:
\[ S = \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} = e + \frac{1}{e} - \cosh(1) \]

### Step 7: Final Calculation
The value of \( S \) can be simplified using known values of \( e \) and \( \cosh(1) \):
\[ S = \frac{e + \frac{1}{e}}{2} \]

### Conclusion
Thus, the final result for the series \( \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} \) is:
\[ S = \frac{e - e^{-1}}{2} \]