Chapter 11 - Series Solutions to Linear Differential Equations - 11.1 A Review of Power Series - Problems - Page 731: 15

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Given: $\sum (n+2)a_{n+1}x^{n}-\sum a_{n}x^n=0\\ \sum x^n[(n+2)a_{n+1}-a_n]=0$ Comparing $x^n$ on both sides: $a_{n+1}=\frac{a_{n}}{n+2}$ It gives: $a_1=\frac{1}{2}a_0\\ a_2=\frac{1}{4}a_1=1^2.\frac{1}{2.3}a_0=\frac{1}{6}a_0\\ a_3=1^3.\frac{a_0}{2.3.4}\\ a_n=1^n.\frac{a_0}{(n+1)!}$ We have: $f(x)=\frac{a_0}{x}\sum \frac{1}{(n+1)!}x^{n+1}$