A differential equation with a singular point at x==1:

Sin[π x] y''[x]  +  Cos[π x] y'[x] + 
        y[x] == 0

Divide both sides of the DE by Sin[π x]:

y''[
          x]  +  (Cos[π x]/Sin[π x]) y'[
            x]  +  (1/Sin[π x]) y[x]  ==  0

Multiply by (x-1)^2 --- we want to examine the singular point at x==1

(x - 1)^2 y''[
            x]  +  ((x - 1)^2 Cos[π x]/Sin[π x]) y'[
            x]  +  ((x - 1)^2/Sin[π x]) y[x]  ==  
      0

The leading term already looks like one from an Euler equation.  We group the middle term so that it does too:  (stuff) (x-1) y'[x].

(x - 1)^2 y''[x] + ((x - 1) Cos[π x]/Sin[π x]) (x - 1) y'[x] + ((x - 1)^2/Sin[π x]) y[x] == 0

The stuff in red above is also known as (x-1) (Q/P) in the parlance of the textbook; confusingly enough, the book also calls this "(x-1)p".  We want to know its behavior near the singular point.

Limit[(x - 1) Cos[π x]/Sin[π x],  
      x -> 1]

(The above limit is (Subscript[p, 0]) (see below) and is aka α.)

The stuff in blue above is also known as (x-1)^2 (R/P), aka "(x-1)^2 q".  We want to know its behavior, too, near the singular point.

Limit[(x - 1)^2/ Sin[π x],  x -> 1]

(The above limit is (Subscript[q, 0]) (see below) and is aka β.)

Thus the Euler-ized version of our DE is:

(x - 1)^2 YE''[x]  +  (1/π) (x - 1) YE'[
            x]  +  0 YE[x]  ==  0

This is a DE whose solution will approximate the behavior of the solution to the original problem near the singular point x==1.

And the indicial equation (which will tell us the exponents near the singularity of the original DE) is:

r (r - 1)  +  (1/π) r  +  0  ==  0

For later convenience, we'll denote the polynomial LHS by "F":

F[r_]  :=  r (r - 1)  +  (1/π) r  +  0

It's roots are:

Solve[ 
      r (r - 1)  +  (1/π) r  +  0  ==  
        0 ]

Note these roots are both real, and do not differ by an integer.  We can use the recurrence relation --- formula (8) on page 274 --- already derived in the book to get the coefficients for our series solution corresponding to each root.  That is, the solution to the Euler-ized equation is:

YE[x_]  :=  C[1] x^0   +   C[2] x^(1 - 1/π)

And the solution to our original problem will be:

y[x_]  :=  
      C[1] x^0 (1 + 
              Sum[a[n] (x - 1)^n,  {n, 1, Infinity}])   +   C[
            2] x^(1 - 1/π) (1 + 
              Sum[b[n] (x - 1)^n,  {n, 1, Infinity}])

This is the Taylor series expansion for the red "(x-1)p":

Series[(x - 1) Cos[π x]/Sin[π x],  {x, 1, 
        5}]

This is the series expansion for the blue "(x-1)^2 q":

Series[(x - 1)^2/Sin[π x],  {x, 1, 5}]

The formulas given in the book use the coefficients of these series.  They set:

red=(x - 1) p  =  
        (Subscript[p, 0])  +  (Subscript[p, 1]) (x - 1)  +  (Subscript[p, 2]) (x - 1)^2  +  
            (Subscript[p, 3]) (x - 1)^3  +   ...

blue = (x - 1)^2 q  =  
        (Subscript[q, 0])  +  (Subscript[q, 1]) (x - 1)  +  (Subscript[q, 2]) (x - 1)^2  +  
            (Subscript[q, 3]) (x - 1)^3  +   ...

To compute,we'll form a table of the coefficient values for the red "(x-1)p":

pcoeffs  =  
      Table[ Limit[ 
          D[ (x - 1) Cos[π x]/Sin[π x],  {x, 
                n} ] / n!,  x -> 1 ],  {n, 0, 5} ]

And the blue "(x-1)^2 q":

qcoeffs  =  
      Table[ Limit[ 
          D[ (x - 1)^2/Sin[π x],  {x, n} ] / n!,  
          x -> 1 ],  {n, 0, 5} ]

The recurrence relation for the coefficients of the solution(s) is: (for n>=1)

F[r + n] c[n]  +  
        Sum[ c[k] ((r + k) pcoeffs[[n - k]] + 
                qcoeffs[[n - k]]),  {k, 0, n - 1} ]   ==  
      0

where "r" is the exponent at the singularity for the solution with coefficients "c[n]".

For the smaller root:

a[0]  =  1;

a[n_] := (-1/F[0+n]) Sum[ a[k] ((0+k) pcoeffs[[n - k]]+qcoeffs[[n - k]]), {k, 0, n - 1} ];

acoeffs  =  Simplify[  Table[ a[n],  {n, 0, 5} ]  ]

N[ acoeffs ]

Note that the above doesn't need to be fudged in any way since the two exponents at the singularity don't differ by an integer.  (If they did, we'd need to further modify the series corresponding to this root.)

For the larger root: (for which the proposed series never needs to be fudged further)

b[0]  =  1;

b[n_] := (-1/F[(1 - 1/π)+n]) Sum[ a[k] (((1 - 1/π)+k) pcoeffs[[n - k]]+qcoeffs[[n - k]]), {k, 0, n - 1} ];

bcoeffs  =  Simplify[  Table[ b[n],  {n, 0, 5} ]  ]

N[ bcoeffs ]

Lastly, we need to ask when the series converge.  Fortunately we don't need to use any tests like those from 408D (comparison, limit comparison, integral, ratio, or root tests).  The theorem on p278-279 tells us that the series for our solution converges within a radius of the center of the series at least as large as the distance to the next singular point.  That is, our DE has singular points at ..., -1, 0, 1, 2, 3, ... .  We centered our series at x=1.  The nearest singular points to our center (other than the one we're sitting on) come at x=0 and x=2 at a distance of one unit away (Abs[2-1]==1 and Abs[0-1]==1).  Thus the radius of convergence for the series is at least one.

The book states this fact in a sort-of backhanded way.  If we look at the series for red "(x-1)p", it will converge within a radius of one of x=1.  If we look at the series for blue "(x-1)^2 q", it will converge within a radius of one of x=1.  The minimum of these two radii is one.