# Incomplete Series Evaluation in Mathematica for Large Values of 𝑛

The evaluation of the series

Series[((1/2 + Sqrt[n])^3 n^(15))/(((7 Sqrt[2])/(3 n^(1/4)) +
C1/n^(3/4) + n^(1/4))^(37)), {n, Infinity, 2}]


Does not complete where C1 is a constant. Hours and no result. However, if I remove the constant then the series

Series[((1/2 + Sqrt[n])^3 n^(15))/(((7 Sqrt[2])/(3 n^(1/4)) + n^(1/4))^(37)),
{n, Infinity, 2}]


Completes in seconds.

What is going on and how do I fix this. This is holding up calculation progress.

Using Mathematical 12.2

Workaround:

\$Version
(*14.0.0 for Microsoft Windows (64-bit) (December 13, 2023)*)

(Series[((1/2 + Sqrt[n])^3 n^B)/(C1/n^(3/4) + (7 Sqrt[2])/(3 n^(1/4)) + n^(1/4))^A, {n, Infinity, 2}, Assumptions -> {A > 0, B > 0, C1 > 0}] //
Normal) /. B -> 15 /. A -> 37 // FullSimplify // Expand

Asymptotic[((1/2 + Sqrt[n])^3  n^B)/(C1/n^(3/4) + (7  Sqrt[2])/(3  n^(1/4)) + n^(1/4))^A, {n,
Infinity, 1}, Assumptions -> {A > 0, B > 0, C1 > 0}] /.
B -> 15 /. A -> 37 // FullSimplify // Expand


In fact, I also don't know why Mathamatica chokes and calculate for long time.

• Well, it is not clear as to why my original statement would not complete even after many hours, I tried your example and it completed in seconds. Commented Jun 23 at 19:08

Workaround:

f[n_] = ((1/2 + Sqrt[n])^3 n^15)/(C1/n^(3/4) + (7 Sqrt[2])/(3 n^(1/4)) + n^(1/4))^37;

Asymptotic[f[n], n -> ∞]
(*    n^(29/4) if C1 ∈ R    *)


Non-fractional expansion in terms of $$m=\sqrt{n}$$:

(Normal[Series[m^(-29/2) f[m^2], {m, ∞, 2}]] /.
m -> Sqrt[n]) n^(29/4) // Expand


$$-37 C_1 n^{25/4}+n^{29/4}-\frac{259}{3} \sqrt{2} n^{27/4}+\frac{3 n^{27/4}}{2}-\frac{259 n^{25/4}}{\sqrt{2}}+\frac{275603 n^{25/4}}{36}$$