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 x^k+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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