13: There is k in the denominator, but also
k + 1, so no known expansion will help.
Thus we need to get rid of these in the denominator. Note that when this
series is differentiated, the k in the denominator disappears, which
is a good start.
Differentiation is
usually handled more easily if you first denote the sum of the given series as
f (x) and then manipulate simultaneously both
sides of this equality.
There is another way to simplify the denominator, using
partial fractions
decomposition.
The third alternative: One can use the first approach, but first adjust the
power to xk+1 and then use differentiation to
remove k + 1 from the denominator. Then there
will be just k left and the logarithmic expansion can be used right
away.
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