Question

Let \[ S = \sum_{k=1}^{\infty} (-1)^{k-1}\frac{1}{k}\left(\frac{1}{4}\right)^k, \quad T = \sum_{k=1}^{\infty} \frac{1}{k}\left(\frac{1}{5}\right)^k. \] Then, which of the following statements is TRUE?

• A. \( S - T = 0 \)
• B. \( 5S - 4T = 0 \)
• C. \( 4S - 5T = 0 \)
• D. \( 16S - 25T = 0 \)

Hint:

Recognize power series forms of \(\ln(1 + x)\) and \(\ln(1 - x)\); alternating signs correspond to \(\ln(1 + x)\), positive to \(-\ln(1 - x)\).

Answer 1

The Correct Option is A

Step 1: Recognize the series type.
Both \(S\) and \(T\) are logarithmic series of the form \[ \sum_{k=1}^{\infty} \frac{r^k}{k} = -\ln(1 - r), \quad |r|<1. \] For alternating signs, \[ \sum_{k=1}^{\infty} (-1)^{k-1}\frac{r^k}{k} = \ln(1 + r). \]
Step 2: Apply to given series.
\[ S = \ln\left(1 + \frac{1}{4}\right) = \ln\left(\frac{5}{4}\right), \] \[ T = -\ln\left(1 - \frac{1}{5}\right) = -\ln\left(\frac{4}{5}\right) = \ln\left(\frac{5}{4}\right). \] Thus, \( S = T \).
Step 3: Verify given options.
If \(S = T\), then \(4S - 5T = 4S - 5S = -S = 0\) (since \(S = T\) implies same ratio). Hence, \(4S - 5T = 0\) is true. Final Answer: \[ \boxed{4S - 5T = 0} \]