# Does this sum converge, and why?

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?

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}]]]]

• 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? Jan 25, 2021 at 20:57
• 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. Jan 25, 2021 at 21:05
• @user64494 Yes, but the 1/m terms get small for large m Jan 25, 2021 at 21:09
• 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 Jan 25, 2021 at 21:12
• @Andreas: You are right. Jan 25, 2021 at 21:16

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]