Let g(x)=∑n=1∞x2n2n⋅ng(x) = \sum_{n=1}^{\infty} \frac{x^{2n}}{2^n \cdot n}g(x)=∑n=1∞2n⋅nx2n. Find g′(x)g'(x)g′(x) by differentiating term-by-term:
∑n=0∞x2n+12n\sum_{n=0}^{\infty} \frac{x^{2n+1}}{2^n}∑n=0∞2nx2n+1
∑n=1∞x2n−12n−1\sum_{n=1}^{\infty} \frac{x^{2n-1}}{2^{n-1}}∑n=1∞2n−1x2n−1
∑n=1∞2x2n−12n\sum_{n=1}^{\infty} \frac{2x^{2n-1}}{2^n}∑n=1∞2n2x2n−1
∑n=0∞x2n2n\sum_{n=0}^{\infty} \frac{x^{2n}}{2^n}∑n=0∞2nx2n