2
$\begingroup$

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

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Commented Jun 23 at 19:08
4
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.