If $0<x<1$, then $\frac{3}{2}x^2 + \frac{5}{3}x^3 + \frac{7}{4}x^4 + \dots$, is equal to :
Show Hint
Break down the general term of a series into simpler components. Often, one part will be a Geometric Progression (GP) and the other will be a standard derivative or integral of a GP (like the log or inverse tangent series).
Step 1: Understanding the Concept:
This is a power series problem. We can rewrite the coefficients to relate the series to standard logarithmic expansions. Step 2: Key Formula or Approach:
The standard expansion is $-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots$. Step 3: Detailed Explanation:
The given series is $S = \sum_{n=2}^{\infty} \frac{2n-1}{n} x^n$.
Let's rewrite the term: $\frac{2n-1}{n} = 2 - \frac{1}{n}$.
\[ S = \sum_{n=2}^{\infty} (2 - 1/n) x^n = 2 \sum_{n=2}^{\infty} x^n - \sum_{n=2}^{\infty} \frac{x^n}{n} \]
Part 1: $2(x^2 + x^3 + x^4 + \dots) = 2 \frac{x^2}{1-x}$.
Part 2: $\frac{x^2}{2} + \frac{x^3}{3} + \dots = [x + \frac{x^2}{2} + \frac{x^3}{3} + \dots] - x = -\ln(1-x) - x$.
So, $S = \frac{2x^2}{1-x} - (-\ln(1-x) - x) = \frac{2x^2}{1-x} + x + \ln(1-x)$.
Combine the fractional terms:
\[ S = \frac{2x^2 + x(1-x)}{1-x} + \ln(1-x) = \frac{2x^2 + x - x^2}{1-x} + \ln(1-x) \]
\[ S = \frac{x^2 + x}{1-x} + \ln(1-x) = x \left( \frac{1 + x}{1 - x} \right) + \ln(1 - x) \] Step 4: Final Answer:
The sum of the series is $x \left( \frac{1 + x}{1 - x} \right) + \log_e(1 - x)$.
