# Compute a Taylor series of a function near zero?

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 Hi everybody!I'm interessed in the behavior near zero of a function depending on a positive parameter"p": Cas2a[\[Phi]_] := Sqrt[-((p + (1 + p) (-1 + (1/(1 + p))^\[Phi]) \[Phi] PolyGamma[ 1, \[Phi]])/(p (1 + p)^2 \[Phi]^2))]; (Phi should be positive too). So, I compute the beginning of the series near 0: Series[Cas2a[\[Phi]], {\[Phi], 0, 1}] Print[Style["Coef order -1 =", Bold, Blue], SeriesCoefficient[Series[Cas2a[\[Phi]], {\[Phi], 0, 1}], -1]]; Print[Style["Coef order 0 =", Bold, Blue], SeriesCoefficient[Series[Cas2a[\[Phi]], {\[Phi], 0, 1}], 0]]; Print[ Style["Coef order 1 =", Bold, Blue], SeriesCoefficient[Series[Cas2a[\[Phi]], {\[Phi], 0, 1}], 1]]; there is a lag in the coefficient! On the contrary, processing a simpler function, there is no problem. Series[Cos[x]/x, {x, 0, 10}] Table[SeriesCoefficient[Series[Cos[x]/x, {x, 0, 10}], i], {i, -1, 3}] Why? What should I do? Thanks in advance, Claude
 Include an assumption that ϕ is positive: Series[Cas2a[ϕ], {ϕ, 0, 1}, Assumptions -> ϕ > 0] //TeXForm  $$\frac{\sqrt{\frac{-p+p \left(-\log \left(\frac{1}{p+1}\right)\right)-\log \left(\frac{1}{p+1}\right)}{p (p+1)^2}}}{\phi }-\frac{(p+1) \log ^2\left(\frac{1}{p+1}\right) \sqrt{\frac{-p+p \left(-\log \left(\frac{1}{p+1}\right)\right)-\log \left(\frac{1}{p+1}\right)}{p (p+1)^2}}}{4 \left(-p-(p+1) \log \left(\frac{1}{p+1}\right)\right)}+\frac{1}{2} \phi \sqrt{\frac{-p+p \left(-\log \left(\frac{1}{p+1}\right)\right)-\log \left(\frac{1}{p+1}\right)}{p (p+1)^2}} \left(-\frac{(p+1)^2 \log ^4\left(\frac{1}{p+1}\right)}{16 \left(-p-(p+1) \log \left(\frac{1}{p+1}\right)\right)^2}-\frac{(p+1) \left(\log ^3\left(\frac{1}{p+1}\right)+\pi ^2 \log \left(\frac{1}{p+1}\right)\right)}{6 \left(-p-(p+1) \log \left(\frac{1}{p+1}\right)\right)}\right)+O\left(\phi ^2\right)$$