`1+(log_e n)^2 /(2!) + (log_e n )^4 / (4!)+...=`

Options

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

Answer

To solve the expression

\[ 1 + \frac{(\log_e n)^2}{2!} + \frac{(\log_e n)^4}{4!} + \ldots \]

we can recognize that this series resembles the series expansion of the exponential function. Let's break it down step by step.

### Step 1: Recognize the Series
The series can be identified as the sum of even powers of \(\log_e n\) divided by their factorials. This is similar to the expansion of \(e^x\) and \(e^{-x}\).

### Step 2: Use the Exponential Series
Recall the series expansion for \(e^x\):

\[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

And for \(e^{-x}\):

\[ e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \ldots \]

### Step 3: Sum the Two Series
If we add these two series, the odd powers will cancel out:

\[ e^x + e^{-x} = 2 \left( 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \right) \]

This means:

\[ \frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots \]

### Step 4: Substitute \(x = \log_e n\)
Now, we can substitute \(x = \log_e n\):

\[ \frac{e^{\log_e n} + e^{-\log_e n}}{2} = 1 + \frac{(\log_e n)^2}{2!} + \frac{(\log_e n)^4}{4!} + \ldots \]

### Step 5: Simplify the Left Side
Now, simplify the left side:

\[ e^{\log_e n} = n \quad \text{and} \quad e^{-\log_e n} = \frac{1}{n} \]

Thus, we have:

\[ \frac{n + \frac{1}{n}}{2} \]

### Step 6: Final Result
Putting it all together, we find:

\[ 1 + \frac{(\log_e n)^2}{2!} + \frac{(\log_e n)^4}{4!} + \ldots = \frac{n + \frac{1}{n}}{2} \]

### Conclusion
The final value of the expression is:

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