Question:

Step 1: Use the property of definite integrals. Since $|x|$ is even and the integrand has even and odd parts, split the sum. Note that $|x| \cdot x^k$ is odd when $k$ is even and even when $k$ is odd... Actually, $|x| \cdot x^k$: for $k$ even, $x^k$ is even, so $|x| x^k$ is even; for $k$ odd, $x^k$ is odd, so $|x| x^k$ is odd. Odd-function integrals over $[-1,1]$ vanish. Step 2: Retain only even-powered terms. The surviving terms are those with even powers of $x$: $$I_n = 2\int_0^1 x\left(1 + \frac{x^2}{2} + \frac{x^4}{4} + \ldots + \frac{x^{2n}}{2n}\right)dx$$ Step 3: Compute $I_2$ explicitly: $$I_2 = 2\int_0^1 x\left(1 + \frac{x^2}{2} + \frac{x^4}{4}\right)dx = 2\int_0^1 \left(x + \frac{x^3}{2} + \frac{x^5}{4}\right)dx$$ $$= 2\left[\frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{24}\right]_0^1 = 2\left(\frac{1}{2} + \frac{1}{8} + \frac{1}{24}\right) = 2 \cdot \frac{12 + 3 + 1}{24} = 2 \cdot \frac{16}{24} = \frac{7}{6}$$ Step 4: Compute $\lim_{n\to\infty} I_n$. As $n \to \infty$: $$I_n = 2\int_0^1 x \sum_{k=0}^{n} \frac{x^{2k}}{2k} \to 2\int_0^1 x \cdot \left(-\frac{1}{2}\ln(1-x^2)\right) dx$$ where we used $\sum_{k=1}^{\infty} \frac{x^{2k}}{2k} = -\frac{1}{2}\ln(1-x^2)$ (plus the $k=0$ term which is 1). Step 5: Evaluate the limit integral: $$\lim_{n\to\infty} I_n = 2\int_0^1 x\left(1 - \frac{1}{2}\ln(1-x^2)\right)dx = 2\left[\frac{1}{2} + \frac{1}{4}\int_0^1 \ln(1-x^2)\cdot(-2x)\,dx\right]$$ Let $u = 1-x^2$, $du = -2x\,dx$: $$= 1 - \frac{1}{2}\int_0^1 \ln u\, du = 1 - \frac{1}{2}[u\ln u - u]_0^1 = 1 - \frac{1}{2}(0 - 1) = 1 + \frac{1}{2} = \frac{3}{2}$$