Sum of the series `log_(e) 2 + ((log_(e)2)^(2))/(2!) + ((log_(e)2)^(3))/(3!) +...=`

Options

A. 3
B. 2
C. 1
D. 0

Answer

To find the sum of the series

\[ S = \log_e 2 + \frac{(\log_e 2)^2}{2!} + \frac{(\log_e 2)^3}{3!} + \ldots \]

we can recognize that this series resembles the Taylor series expansion for the exponential function \( e^x \), which is given by:

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

### Step 1: Identify the series
We can rewrite our series \( S \) as:

\[ S = \sum_{n=1}^{\infty} \frac{(\log_e 2)^n}{n!} \]

This is similar to the series for \( e^x \), but it starts from \( n=1 \).

### Step 2: Relate it to the exponential function
Notice that the series for \( e^{\log_e 2} \) is:

\[ e^{\log_e 2} = 1 + \frac{(\log_e 2)^1}{1!} + \frac{(\log_e 2)^2}{2!} + \frac{(\log_e 2)^3}{3!} + \ldots \]

This means:

\[ e^{\log_e 2} = 2 \]

### Step 3: Adjust for the starting index
Since our series \( S \) starts from \( n=1 \), we can express it in terms of the full series:

\[ S = e^{\log_e 2} - 1 \]

### Step 4: Substitute the value
Substituting the value from the exponential function:

\[ S = 2 - 1 = 1 \]

### Conclusion
Thus, the sum of the series is:

\[ \boxed{1} \]