I would like the Taylor Expansion of $$ \frac{\log(u-1) + 1}{u} $$ about the point $u=2.5$ in closed form. The best I've gotten so far is Taylor expand $\log(u-1) + 1$ and $1/u$ separately about $u=2.5$ and then I get a product of two sums. $$ \frac{\log(u-1)+1}{u} = \left( - \sum_{n=0}^\infty (-2/5)^{n+1}(u-5/2)^n \right) \left( 1 + \log(3/2) - \sum_{k=1}^\infty (-2/3)^k \frac{(u-5/2)^k}{k} \right) $$ Reassuringly, Mathematica Simplifies this to the equation given, so I know that up to this point I haven't made a mistake.

Simplify[-Sum[(-2/5)^(k + 1)*(u - 5/2)^k, {k, 0, \[Infinity]}]*
    (1+Log[3/2] - Sum[(-2/3)^k*(u - 5/2)^k/k, {k, 1, \[Infinity]}])]


But I would like Mathematica to give me a closed form for $a_k$, where $$ \frac{\log(u-1)+1}{u} = \sum_{k=0}^\infty a_k (u-5/2)^k $$ and that's where I'm stuck.

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    $\begingroup$ Have you seen SeriesCoefficient[]? $\endgroup$ – J. M.'s technical difficulties Apr 6 '17 at 9:49
  • $\begingroup$ Yeah, Mathematica thinks its really clever and reduces the product of two sums immediately to $\frac{1 + \log(-1 + u)}{u}$, so trying SeriesCoefficient on the above product of series doesn't work. $\endgroup$ – J. Ashford Apr 6 '17 at 10:00
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    $\begingroup$ I was thinking that since you're expanding at a different point, then you'd use SeriesCoefficient[fun, {u, 5/2, n}]... $\endgroup$ – J. M.'s technical difficulties Apr 6 '17 at 10:02
  • $\begingroup$ Right, I think that's sufficient to get to the answer I wanted, thanks! $\endgroup$ – J. Ashford Apr 6 '17 at 10:21

As noted, this is as simple as evaluating

SeriesCoefficient[(1 + Log[u - 1])/u, {u, 5/2, n}]

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