One way to do it is application of series of rules to the expression, just to "teach" Mathematica of how to transform the expression. I will break the chain of transformations into steps to make it better visible.
These are your functions:  

     expr1 = (x - x0)^α  Sum[a[n] (x - x0)^n, {n, 0, Infinity}]
    expr2 = D[expr1, x]
and here are the results:

![enter image description here][1]

Let us do the first transformation:

    expr3 = expr2 /. u_*\!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]v_\) -> \!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\((u* v)\)\)
this is the result:

![enter image description here][2]

The second transformation:

     expr4 = expr3 /. (x - x0)^m_ \!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(s_*\ 
    \*SuperscriptBox[\((x - x0)\), \(q_\)]\)\) -> \!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(s*\ 
    \*SuperscriptBox[\((x - x0)\), \(q + m\)]\)\)

yielding

![enter image description here][3]    

and the last one:

   

     expr5 = expr4 /. \!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(u1_*
          v_\)\) + \!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(u2_*
          v_\)\) -> \!\(
    \*UnderoverscriptBox[\(\[Sum]\), \(n = 
          0\), \(\[Infinity]\)]\(\((u1 + u2)\)*v\)\)

gives the result:

![enter image description here][4]


  [1]: https://i.sstatic.net/bIQk6.gif
  [2]: https://i.sstatic.net/kqYFl.gif
  [3]: https://i.sstatic.net/NqDDO.gif
  [4]: https://i.sstatic.net/e8QkC.gif