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Mathematica says the following sum

    Sum[(mm Gamma[mm])/
  Gamma[-(1/2) + mm] - (mm^(3/2) - (3 Sqrt[mm])/8 - (7 Sqrt[1/mm])/
    128), {mm, 1, \[Infinity]}]

converges and equals

1/(5 Sqrt[\[Pi]]) - Zeta[-(3/2)] + 3/8 Zeta[-(1/2)] + 7/128 Zeta[1/2]

This seems to make sense, bc the large $mm$ limit

Series[(mm Gamma[mm])/Gamma[-(1/2) + mm], {mm, \[Infinity], 1}]

gives

mm^(3/2) - (3 Sqrt[mm])/8 - (7 Sqrt[1/mm])/128+O(m^{-3/2})

which is cancelled by construction in the above convergent sum. But now lets seperately sum each term in the sum to $m$, and look at large $m$:

 Series[Sum[(mm Gamma[mm])/
  Gamma[-(1/2) + mm], {mm, 1, m}], {m, \[Infinity], 0}]

gives

(2 m^(5/2))/5 + m^(3/2)/4 - (11 Sqrt[m])/64 + 1/(5 Sqrt[\[Pi]])+O(m^{-1/2})

while

 Series[Sum[
   mm^(3/2) - (3 Sqrt[mm])/8 - (7 Sqrt[1/mm])/128, {mm, 1, 
    m}], {m, \[Infinity], 0}] // Simplify

gives

  (2 m^(5/2))/5 + m^(3/2)/4 - (19 Sqrt[m])/64 + (Zeta[-(3/2)] - 3/8 Zeta[-(1/2)] - 7/128 Zeta[1/2])+O(m^{-1/2})

Note that the $m$-independent parts equal the result of the convergent sum above, but the $m^{1/2}$ terms are different. How is this possible? In particular, how could the sum in the beginning be convergent, if each term has a different tail?

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2 Answers 2

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If you develop both terms up to second order the m^(1/2) is no longer there:

Expand[FullSimplify[
Normal[Series[Sum[(mm*Gamma[mm])/Gamma[-(1/2) + mm], {mm, 1, m}], 
       {m, Infinity, 2}]]]] - 
Expand[
FullSimplify[
Normal[Series[Sum[mm^(3/2) - 7/(128*Sqrt[mm]) - (3*Sqrt[mm])/8, 
         {mm, 1, m}], {m, Infinity, 2}]]]]
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  • $\begingroup$ hm, thats weird. How come we needed to expand to second order, in order to get the correct results at a subleading order? this looks like a bug... how do i know if this issue wont show up elsewhere? $\endgroup$
    – esches
    Jan 25, 2021 at 20:57
  • $\begingroup$ No, it is not so for me: Series[Sum[(mm Gamma[mm])/Gamma[-(1/2) + mm], {mm, 1, m}], {m, \[Infinity], 2}] - Series[Sum[ mm^(3/2) - (3 Sqrt[mm])/8 - (7 Sqrt[1/mm])/128, {mm, 1, m}], {m, \[Infinity], 2}] // Simplify produces $$\left(-\zeta \left(-\frac{3}{2}\right)+\frac{3 \zeta \left(-\frac{1}{2}\right)}{8}+\frac{7 \zeta \left(\frac{1}{2}\right)}{128}+\frac{1}{5 \sqrt{\pi }}\right)+\frac{9 \sqrt{\frac{1}{m}}}{512}-\frac{1247 \left(\frac{1}{m}\right)^{3/2}}{245760}+O\left(\left(\frac{1}{m}\right)^{5/2}\right) $$ in 12.2. $\endgroup$
    – user64494
    Jan 25, 2021 at 21:05
  • $\begingroup$ @user64494 Yes, but the 1/m terms get small for large m $\endgroup$
    – Andreas
    Jan 25, 2021 at 21:09
  • $\begingroup$ it seems the asymptotic series of the square root expression 'converges' slower than the gamma term. It works also when you develop only the second part to second order $\endgroup$
    – Andreas
    Jan 25, 2021 at 21:12
  • $\begingroup$ @Andreas: You are right. $\endgroup$
    – user64494
    Jan 25, 2021 at 21:16
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According to AsymptoticSum, the leading term for the series is constant at infinity:

AsymptoticSum[
    (mm Gamma[mm])/ Gamma[-(1/2) + mm] - (mm^(3/2) - (3 Sqrt[mm])/8 - (7 Sqrt[1/mm])/128), 
    {mm, 1, n},
    n -> \[Infinity]
]

1/(5 Sqrt[[Pi]]) - Zeta[-(3/2)] + 3/8 Zeta[-(1/2)] + 7/128 Zeta[1/2]

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