I want to expand the argument of Exp[ Sqrt[1+x]/x ] in powers of x around x = 0.

Series[  Exp[ Sqrt[1+x]/x ] ,{x,0,3}] 

does not work as there is an essential singularity.

The next best thing is to do

Exp[ Series[ Sqrt[1+x]/x ,{x,0,3}]  ] 

But I want to do this outside the Exp function as there may be many such terms all added up...

  • 1
    $\begingroup$ You suggest this example is not entirely representative, so I don't know whether this will work in all your cases: Exp[Sqrt[1 + x]/x] /. Exp[u_] :> Exp[Series[u, {x, 0, 3}]] $\endgroup$
    – Michael E2
    Commented May 20, 2017 at 13:12
  • $\begingroup$ Mr. Michael E2 , this will not do as I want to do this outside the argument of Exp function not inside but the effect has to be same as expanding the argument inside the argument of Exp. $\endgroup$ Commented May 21, 2017 at 13:32
  • 1
    $\begingroup$ Then I do not understand what output you want. Please update the question with the formula you want as the output for the expansion of Exp[ Sqrt[1+x]/x ]. $\endgroup$
    – Michael E2
    Commented May 21, 2017 at 14:08
  • $\begingroup$ Imagine I want to expand the exponents of each of these terms (below) up to order x^3 . (x^2 + Cos[x]) Exp[Sqrt[1 + x]/x] + $\endgroup$ Commented May 22, 2017 at 16:23
  • $\begingroup$ What would be the output you would expect? In particular what are the terms up to order x^3 of Exp[Sqrt[1 + x]/x] that you would expect to see? $\endgroup$
    – Michael E2
    Commented May 22, 2017 at 17:00

1 Answer 1


We can use the identity $e^{x+y} = e^x e^y$ to split the argment of Exp into a finite part and an infinite part. Here is some code to extract the finite and infinite parts of a series:

infinitePart[s:HoldPattern @ SeriesData[x_, x0_, __, inc_]] := Normal[
    s + SeriesData[x, x0, {}, 0, 0, inc]
finitePart[s_SeriesData] := s - infinitePart[s]

Now, making use of the identity:

    Series[Exp[Sqrt[1+x]/x], {x, 0, 3}],
    Exp[s_SeriesData] :> Exp[infinitePart[s]] Exp[finitePart[s]
] //TeXForm

$e^{\frac{1}{x}} \left(\sqrt{e}-\frac{\sqrt{e} x}{8}+\frac{9 \sqrt{e} x^2}{128}-\frac{145 \sqrt{e} x^3}{3072}+O\left(x^4\right)\right)$

we obtain the form I think you're looking for.


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