The `3^(rd)` term of `log_(e) 2` is

Answer

To find the third term of \( \log_{e} 2 \), we can use the Taylor series expansion for \( \log_{e}(1+x) \). Here are the steps to derive the solution:

### Step 1: Rewrite the logarithm
We can express \( \log_{e} 2 \) as:
\[ \log_{e} 2 = \log_{e}(1 + 1) \]

### Step 2: Use the Taylor series expansion
The Taylor series expansion for \( \log_{e}(1+x) \) around \( x=0 \) is given by:
\[ \log_{e}(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
In this case, we set \( x = 1 \).

### Step 3: Substitute \( x = 1 \) into the series
Substituting \( x = 1 \) into the series gives:
\[ \log_{e}(1+1) = 1 - \frac{1^2}{2} + \frac{1^3}{3} - \frac{1^4}{4} + \ldots \]
This simplifies to:
\[ \log_{e}(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \]

### Step 4: Identify the terms
The series can be written as:
- First term: \( 1 \)
- Second term: \( -\frac{1}{2} \)
- Third term: \( \frac{1}{3} \)
- Fourth term: \( -\frac{1}{4} \)

### Step 5: Find the third term
From the series, we see that the third term is:
\[ \frac{1}{3} \]

### Conclusion
Thus, the third term of \( \log_{e} 2 \) is:
\[ \frac{1}{3} \]

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