If `|a| lt 1, b = sum_(k=1)^(oo) (a^(k))/(k) rArr a=`

Options

A. `sum_(k=1)^(oo)((-1)^(k)b^(k))/(k)`
B. `sum_(k=1)^(oo)((-1)^(k-1)b^(k))/(k!)`
C. `sum_(k=1)^(oo)((-1)^(k)b^(k))/((k-1)!)`
D. `sum_(k=1)^(oo)((-1)^(k-1)b^(k))/((k+1)!)`

Answer

To find \( a \) in terms of \( b \) given that \( b = \sum_{k=1}^{\infty} \frac{a^k}{k} \) and \( |a| < 1 \), we can follow these steps:

### Step 1: Recognize the Series
The series \( b = \sum_{k=1}^{\infty} \frac{a^k}{k} \) is known to be the Taylor series expansion for \( -\ln(1 - a) \) when \( |a| < 1 \).

### Step 2: Write the Equation
From the Taylor series, we have:
\[ b = -\ln(1 - a) \]

### Step 3: Solve for \( a \)
To express \( a \) in terms of \( b \), we first isolate \( \ln(1 - a) \):
\[ \ln(1 - a) = -b \]
Now, exponentiate both sides to remove the logarithm:
\[ 1 - a = e^{-b} \]

### Step 4: Rearrange to Find \( a \)
Now, we can solve for \( a \):
\[ a = 1 - e^{-b} \]

### Final Result
Thus, we have expressed \( a \) in terms of \( b \):
\[ a = 1 - e^{-b} \]