Coefficient of `n^(-r)` in the expansion of `log_(10)((n)/(n-1))` is

Options

A. `(1)/("rlog"_(e )10)`
B. `-(1)/("rlog"_(e )10)`
C. `-(1)/("r!log"_(6)10)`
D. `(1)/("log"_(6)10)`

Answer

To find the coefficient of \( n^{-r} \) in the expansion of \( \log_{10}\left(\frac{n}{n-1}\right) \), we can follow these steps:

### Step 1: Rewrite the logarithmic expression
We start with the expression:
\[ \log_{10}\left(\frac{n}{n-1}\right) \]
Using the change of base formula, we can rewrite this as:
\[ \log_{10}\left(\frac{n}{n-1}\right) = \frac{\ln\left(\frac{n}{n-1}\right)}{\ln(10)} \]

### Step 2: Simplify the logarithm
Next, we simplify the natural logarithm:
\[ \ln\left(\frac{n}{n-1}\right) = \ln(n) - \ln(n-1) \]
Thus, we have:
\[ \log_{10}\left(\frac{n}{n-1}\right) = \frac{\ln(n) - \ln(n-1)}{\ln(10)} \]

### Step 3: Expand \(\ln(n-1)\)
We can use the Taylor expansion for \(\ln(n-1)\) around \(n\):
\[ \ln(n-1) = \ln\left(n\left(1 - \frac{1}{n}\right)\right) = \ln(n) + \ln\left(1 - \frac{1}{n}\right) \]
Using the series expansion for \(\ln(1-x)\):
\[ \ln(1 - x) = -\sum_{r=1}^{\infty} \frac{x^r}{r} \quad \text{for } |x| < 1 \]
we can substitute \(x = \frac{1}{n}\):
\[ \ln\left(1 - \frac{1}{n}\right) = -\sum_{r=1}^{\infty} \frac{1}{rn^r} \]
Thus,
\[ \ln(n-1) = \ln(n) - \sum_{r=1}^{\infty} \frac{1}{rn^r} \]

### Step 4: Substitute back into the logarithm
Now substituting this back into our expression:
\[ \log_{10}\left(\frac{n}{n-1}\right) = \frac{\ln(n) - \left(\ln(n) - \sum_{r=1}^{\infty} \frac{1}{rn^r}\right)}{\ln(10)} \]
This simplifies to:
\[ \log_{10}\left(\frac{n}{n-1}\right) = \frac{\sum_{r=1}^{\infty} \frac{1}{rn^r}}{\ln(10)} \]

### Step 5: Identify the coefficient of \(n^{-r}\)
From the series expansion, we can see that the coefficient of \(n^{-r}\) in the series \(\sum_{r=1}^{\infty} \frac{1}{rn^r}\) is simply \(\frac{1}{r}\). Therefore, the coefficient of \(n^{-r}\) in the expansion of \(\log_{10}\left(\frac{n}{n-1}\right)\) is:
\[ \frac{1}{r \ln(10)} \]

### Final Answer
Thus, the coefficient of \(n^{-r}\) in the expansion of \(\log_{10}\left(\frac{n}{n-1}\right)\) is:
\[ \frac{1}{r \ln(10)} \]