The sum of the series ` 1 +(1)/(4.2!)+ (1)/(16.4!) + (1)/(64.6!) + ...oo` is

Options

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

Answer

To find the sum of the series \( S = 1 + \frac{1}{4 \cdot 2!} + \frac{1}{16 \cdot 4!} + \frac{1}{64 \cdot 6!} + \ldots \), we can follow these steps:

### Step 1: Rewrite the series
We can express the terms in the series in a more manageable form. Notice that the denominators can be rewritten as:
\[ S = 1 + \frac{1}{2^2 \cdot 2!} + \frac{1}{2^4 \cdot 4!} + \frac{1}{2^6 \cdot 6!} + \ldots \]
This gives us:
\[ S = 1 + \frac{1}{2^2 \cdot 2!} + \frac{1}{2^4 \cdot 4!} + \frac{1}{2^6 \cdot 6!} + \ldots \]

### Step 2: Identify the pattern
We see that the series can be expressed as:
\[ S = 1 + \sum_{n=1}^{\infty} \frac{1}{(2n)!(2^n)} \]
This is because \( 4^n = (2^2)^n = 2^{2n} \).

### Step 3: Use the exponential series
Recall the Taylor series expansion for \( e^x \):
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]
From this, we can derive that:
\[ \cosh(x) = \frac{e^x + e^{-x}}{2} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \]
If we substitute \( x = \frac{1}{2} \):
\[ \cosh\left(\frac{1}{2}\right) = \sum_{n=0}^{\infty} \frac{(1/2)^{2n}}{(2n)!} \]

### Step 4: Relate to our series
Notice that our series resembles the series for \( \cosh \):
\[ S = 1 + \sum_{n=1}^{\infty} \frac{(1/2)^{2n}}{(2n)!} = 1 + \left(\cosh\left(\frac{1}{2}\right) - 1\right) = \cosh\left(\frac{1}{2}\right) \]

### Step 5: Calculate \( \cosh\left(\frac{1}{2}\right) \)
We can calculate \( \cosh\left(\frac{1}{2}\right) \) using the definition:
\[ \cosh\left(\frac{1}{2}\right) = \frac{e^{1/2} + e^{-1/2}}{2} \]

### Step 6: Final expression for \( S \)
Thus, we have:
\[ S = \cosh\left(\frac{1}{2}\right) = \frac{e^{1/2} + e^{-1/2}}{2} \]
This can be simplified further:
\[ S = \frac{e^{1/2} + \frac{1}{e^{1/2}}}{2} = \frac{e^{1/2} + \frac{1}{e^{1/2}}}{2} = \frac{e^{1/2} + e^{-1/2}}{2} = \frac{e + 1}{2\sqrt{e}} \]

### Final Answer
Thus, the sum of the series is:
\[ S = \frac{e + 1}{2\sqrt{e}} \]