# Computing products of Two Taylor Expansions

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]}])]

Out=(1+Log[-1+u])/u


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.

• Have you seen SeriesCoefficient[]? Apr 6, 2017 at 9:49
• 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. Apr 6, 2017 at 10:00
• I was thinking that since you're expanding at a different point, then you'd use SeriesCoefficient[fun, {u, 5/2, n}]... Apr 6, 2017 at 10:02
• Right, I think that's sufficient to get to the answer I wanted, thanks! Apr 6, 2017 at 10:21

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