The confluent hypergeometric series with parameter $(a,b)$ is
$$F(a,b,x)=\sum_{k=0}^{\infty}\frac{(a)_kx^k}{(b)_kk!},$$
where $(a)_n=a(a+1)\cdots (a+n-1)$ is the Pochhammer symbol. Prove that $F$ satisfies the differential equation
$$xf''(x)+(b-x)f'(x)-af(x)=0.$$
I tried to substitute the series into the differential equation, but then I dont see how the terms can cancel and vanish, someone please helps, thank you very much.
$$F = \sum_{k=0}^{\infty}\frac{(a)_kx^k}{(b)_kk!}$$ $$F' = \sum_{k=0}^{\infty}\frac{(a)_{k+1}x^k}{(b)_{k+1}k!}$$ $$F'' = \sum_{k=0}^{\infty}\frac{(a)_{k+2}x^k}{(b)_{k+2}k!}$$
Hence : \begin{eqnarray*} ODE &=& \sum_{k=0}^{\infty}\frac{(a)_{k+2}x^{k+1}}{(b)_{k+2}k!} + b\sum_{k=0}^{\infty}\frac{(a)_{k+1}x^k}{(b)_{k+1}k!} - \sum_{k=0}^{\infty}\frac{(a)_{k+1}x^{k+1}}{(b)_{k+1}k!} - a\sum_{k=0}^{\infty}\frac{(a)_{k}x^k}{(b)_{k}k!}\\ &=&\sum_{k=0}^{\infty}\frac{x^{k+1}}{k!}\bigg[\frac{(a)_{k+2}}{(b)_{k+2}} - \frac{(a)_{k+1}}{(b)_{k+1}}\bigg] + \sum_{k=0}^{\infty}\frac{x^{k}}{k!}\bigg[b\frac{(a)_{k+1}}{(b)_{k+1}} - a\frac{(a)_{k}}{(b)_{k}}\bigg]\\ &=&\sum_{k=0}^{\infty}\frac{x^{k+1}}{k!}\bigg[\frac{(a)_{k+1}}{(b)_{k+1}}\frac{a+k+1}{b+k+1} - \frac{(a)_{k+1}}{(b)_{k+1}}\bigg] + \sum_{k=0}^{\infty}\frac{x^{k}}{k!}\bigg[\frac{(a)_{k}}{(b)_{k}}\frac{b(a+k)}{b+k} - a\frac{(a)_{k}}{(b)_{k}}\bigg]\\ &=&\sum_{k=0}^{\infty}\frac{x^{k+1}}{k!}\bigg[\frac{(a)_{k+1}}{(b)_{k+1}}\frac{a - b}{b+k+1}\bigg] + \sum_{k=0}^{\infty}\frac{x^{k}}{k!}\bigg[\frac{(a)_{k}}{(b)_{k}}\frac{k(b-a)}{b+k}\bigg]\\ &=&\sum_{k=0}^{\infty}\frac{x^{k+1}}{k!}\bigg[\frac{(a)_{k+1}}{(b)_{k+1}}\frac{a - b}{b+k+1}\bigg] + \sum_{k=1}^{\infty}\frac{x^{k}}{k!}\bigg[\frac{(a)_{k}}{(b)_{k}}\frac{k(b-a)}{b+k}\bigg]\\ &=&\sum_{k=0}^{\infty}\frac{x^{k+1}}{k!}\bigg[\frac{(a)_{k+1}}{(b)_{k+1}}\frac{a - b}{b+k+1}\bigg] + \sum_{k=1}^{\infty}\frac{x^{k}}{(k-1)!}\bigg[\frac{(a)_{k}}{(b)_{k}}\frac{b-a}{b+k}\bigg]\\ &=&\sum_{k=0}^{\infty}\frac{x^{k+1}}{k!}\bigg[\frac{(a)_{k+1}}{(b)_{k+1}}\frac{a - b}{b+k+1}\bigg] + \sum_{s=0}^{\infty}\frac{x^{s+1}}{s!}\bigg[\frac{(a)_{s+1}}{(b)_{s+1}}\frac{b-a}{b+s+1}\bigg]\\ &=&0 \end{eqnarray*}
I went slowly. Each step is a small calculation so you should be able to understand it all. The last step is zero because they are the same but with opposite signs $a-b$ in the first and $b-a$ in the second. Also I let $s+1 = k$ in the end.