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.

  • 1
    $\begingroup$ Have you seen SeriesCoefficient[]? $\endgroup$ Commented Apr 6, 2017 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
    Commented Apr 6, 2017 at 10:00
  • 2
    $\begingroup$ I was thinking that since you're expanding at a different point, then you'd use SeriesCoefficient[fun, {u, 5/2, n}]... $\endgroup$ Commented Apr 6, 2017 at 10:02
  • $\begingroup$ Right, I think that's sufficient to get to the answer I wanted, thanks! $\endgroup$
    – J. Ashford
    Commented Apr 6, 2017 at 10:21

1 Answer 1


As noted, this is as simple as evaluating

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.