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

Options

A. `(1)/(r log 10)`
B. `(1)/(r log 10)`
C. `(1)/((r!)log10)`
D. `-(1)/((r!) log10)`

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 logarithm
We start with the expression:
\[ \log_{10} \left( \frac{n}{n-1} \right) \]
Using the change of base formula for logarithms, we can express this as:
\[ \log_{10} \left( \frac{n}{n-1} \right) = \frac{\log_e \left( \frac{n}{n-1} \right)}{\log_e 10} \]

### Step 2: Simplify the logarithm
Next, we simplify \( \log_e \left( \frac{n}{n-1} \right) \):
\[ \log_e \left( \frac{n}{n-1} \right) = \log_e n - \log_e (n-1) \]
Now we can write:
\[ \log_{10} \left( \frac{n}{n-1} \right) = \frac{\log_e n - \log_e (n-1)}{\log_e 10} \]

### Step 3: Expand \( \log_e (n-1) \)
We can use the Taylor series expansion for \( \log_e (1-x) \) around \( x = 0 \):
\[ \log_e (1-x) = -\left( x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots \right) \]
In our case, we can write \( n-1 \) as \( n(1 - \frac{1}{n}) \):
\[ \log_e (n-1) = \log_e n + \log_e \left(1 - \frac{1}{n}\right) \]
Thus, we have:
\[ \log_e (n-1) = \log_e n - \left( \frac{1}{n} + \frac{1}{2n^2} + \frac{1}{3n^3} + \ldots \right) \]

### Step 4: Substitute back into the logarithm
Substituting this back, we get:
\[ \log_{10} \left( \frac{n}{n-1} \right) = \frac{\log_e n - \left( \log_e n - \left( \frac{1}{n} + \frac{1}{2n^2} + \frac{1}{3n^3} + \ldots \right) \right)}{\log_e 10} \]
This simplifies to:
\[ \log_{10} \left( \frac{n}{n-1} \right) = \frac{\frac{1}{n} + \frac{1}{2n^2} + \frac{1}{3n^3} + \ldots}{\log_e 10} \]

### Step 5: Identify the coefficient of \( n^{-r} \)
From the expansion, the general term can be expressed as:
\[ \frac{1}{k n^k} \quad \text{for } k = 1, 2, 3, \ldots \]
Thus, the coefficient of \( n^{-r} \) in this expansion is:
\[ \frac{1}{r \log_e 10} \]

### Conclusion
The coefficient of \( n^{-r} \) in the expansion of \( \log_{10} \left( \frac{n}{n-1} \right) \) is:
\[ \frac{1}{r \log_e 10} \]