Question

For \(-1\le x\le1\), if f (x) is the sum of the convergent power series \(x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+...+\frac{x^n}{n^2}+...\) then \(f(\frac{1}{2})\) is equal to

• A. \(\displaystyle\int\limits^{\frac{1}{2}}_{0}\frac{In(1-t)}{t}dt.\)
• B. \(-\displaystyle\int\limits^{\frac{1}{2}}_{0}\frac{In(1-t)}{t}dt.\)
• C. \(\displaystyle\int\limits^{\frac{1}{2}}_{0}tIn(1+t)dt.\)
• D. \(\displaystyle\int\limits^{\frac{1}{2}}_{0}tIn(1-t)dt.\)

Answer 1

The Correct Option is B

To solve the given problem, we need to determine the sum of the power series:

\(f(x) = x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+\ldots\) for \(-1 \le x \le 1\).

We are particularly interested in finding \(f\left(\frac{1}{2}\right)\).

Let's analyze this power series. This series resembles a form of a generalized logarithmic series. By transforming the terms, we can relate the given series with the integral definition of the natural logarithm.

Let's recall the integral formula for the logarithm:

\(- \ln(1-t) = \int_{0}^{t} \frac{x}{1-x} \, dx\).

Given this identity, we can conclude that the integral form of the natural log function can be used to derive the series. Here, we see an analogous relationship with the series which leads us to:

\( f(x) = - \int_{0}^{x} \frac{\ln(1-t)}{t} \, dt\).

Now, substituting \(x = \frac{1}{2}\) in the expression obtained:

\( f\left(\frac{1}{2}\right) = - \int_{0}^{\frac{1}{2}} \frac{\ln(1-t)}{t} \, dt\).

Hence, the correct answer for \(f\left(\frac{1}{2}\right)\) is:

\(-\int\limits^{\frac{1}{2}}_{0}\frac{\ln(1-t)}{t}\,dt \).

Thus, the correct option is the second one:

\(-\int\limits^{\frac{1}{2}}_{0}\frac{\ln(1-t)}{t}\,dt \).