3
$\begingroup$

I wanted to compute the series defined by

$$\sum_{k=1}^\infty\frac{(-1)^{k+1}}k x^\underline k$$

where $x^\underline k:=\prod_{j=0}^{k-1}(x-j)$ is a falling factorial. Thus I write

Sum[(-1)^(k + 1)/k FactorialPower[x, k], {k, 1, Infinity}]

but Mathematica says (without more explanation) that "the series does not converges". Thus I thought that this can be certainly true for some values of $x$ so I tried to compute this time

Sum[(-1)^(k + 1)/k FactorialPower[2, k], {k, 1, Infinity}]

what is a finite sum because FactorialPower[2,k] is zero when $k>2$. However Mathematica also said that "the series doe not converges". So, what is going on? This is a bug or there is some technicality that Im not seeing?

Improve this question
$\endgroup$
  • $\begingroup$ Replacing Sum by Table such as in Table[(-1)^(k + 1)/k FactorialPower[2, k], {k, 1, 1000}] is rather convincing that the series converges... Presumably, MMA does not detect that FactorialPower should simplify. $\endgroup$ – anderstood Jan 4 '18 at 19:09
  • $\begingroup$ If you define manually factpow[x_, k_] := Product[(x - j), {j, 0, k - 1}], factpow[x, k] simplifies to (1 - k + x) Pochhammer[2 - k + x, -1 + k], but still the sum does not converge. $\endgroup$ – anderstood Jan 4 '18 at 19:11
  • $\begingroup$ NSum[(-1)^(k + 1)/k *FactorialPower[2, k], {k, 1, Infinity}] returns ComplexInfinity... $\endgroup$ – anderstood Jan 4 '18 at 19:13
  • $\begingroup$ Do you not want to restrict $x$ to integers? $\endgroup$ – JimB Jan 4 '18 at 19:14
  • 1
    $\begingroup$ If $x\ge 1$ and $x$ is an integer, why not replace $\infty$ with $x$ ? I know that doesn't explain why using $\infty$ doesn't work but it does give an answer. $\endgroup$ – JimB Jan 4 '18 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.