The coefficient of n^{-r} in the expansion of | vedclass

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(A) We have $\log_{10}\left(\frac{n}{n-1}\right) = \log_e\left(\frac{n}{n-1}\right) \cdot \log_{10}e$. $= -\log_e\left(\frac{n-1}{n}\right) \cdot \log

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$\frac{1}{r \log_e 10}$

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(A) We have $\log_{10}\left(\frac{n}{n-1}\right) = \log_e\left(\frac{n}{n-1}\right) \cdot \log_{10}e$. $= -\log_e\left(\frac{n-1}{n}\right) \cdot \log_{10}e = -\log_{10}e \cdot \log_e\left(1 - \frac{1}{n}\right)$. Using the logarithmic series expansion $\log_e(1-x) = -\sum_{r=1}^{\infty} \frac{x^r}{r}$ for $|x| $= -\log_{10}e \cdot \left(-\sum_{r=1}^{\infty} \frac{(1/n)^r}{r}\right) = \log_{10}e \cdot \sum_{r=1}^{\infty} \frac{n^{-r}}{r}$. Thus,the coefficient of $n^{-r}$ is $\frac{\log_{10}e}{r} = \frac{1}{r \log_e 10}$.

The coefficient of $n^{-r}$ in the expansion of $\log_{10}\left(\frac{n}{n-1}\right)$ is

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