5
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Evaluate this limit analytically

Limit[Sum[Sqrt[1 + i^2/n^2]/n, {i, n}], n -> Infinity]

Mathematica 11.3 gives 0.

But using numerical calculations

Needs["NumericalCalculus`"];
NLimit[Sum[Sqrt[1 + i^2/n^2]/n, {i, n}], n -> Infinity]

gives 1.1477936343307347.

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5
  • $\begingroup$ What is the value of i? $\endgroup$
    – Anixx
    Jun 22, 2018 at 7:37
  • 1
    $\begingroup$ Yes, possible a bug. It should be returns unevaluated. $\endgroup$ Jun 22, 2018 at 8:52
  • 1
    $\begingroup$ Maple 2018.1 outputs $ 1/2\,\sqrt {2}+1/2\,\ln \left( 1+\sqrt {2} \right) .$ This is a certain definite integral as the limit of its integral sums. $\endgroup$
    – user64494
    Jun 22, 2018 at 9:38
  • $\begingroup$ For an approximation: Limit[Sum[ Series[Sqrt[1 + i^2/n^2]/n, {n, Infinity, 125}] // Normal, {i, n}], n -> Infinity] // N $\endgroup$
    – Bob Hanlon
    Jun 22, 2018 at 21:37
  • $\begingroup$ Now returns unevaluated in 13.3.0. $\endgroup$ Sep 2, 2023 at 5:00

1 Answer 1

3
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This is not the answer to your question, but how to find symbolic solution to the limit.


Using identity:

$$\sum _{k=0}^{\infty } -\frac{\Gamma \left(k-\frac{1}{2}\right) (1-x)^k}{\left(2 \sqrt{\pi }\right) k!}=\sqrt{x}$$

So:

func = -1/(2 Sqrt[Pi])*Gamma[k - 1/2]/(k! n)*(1 - x)^k /. 
x -> (1 + i^2/n^2) // PowerExpand

Sum[Limit[Sum[func, {i, n}], n -> Infinity, Assumptions -> k >= 0], {k, 0, Infinity}]

(* 1/2 (Sqrt[2] + ArcSinh[1]) *)

N[%,20]
(* 1.1477935746963190370 *)
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  • 1
    $\begingroup$ Nice idea.Is it possible to evaluate Limit[Sum[Sqrt[(1 - 4*i + 4*i^2 - 4*n + 8*i*n + 5*n^2)/n^4], {i, n}], n -> ∞] in this way? The symbolic solution should be 1/4 (-2 Sqrt[5]+4 Sqrt[17]-ArcSinh[2]+ArcSinh[4]). $\endgroup$
    – matrix42
    Jun 22, 2018 at 10:30
  • $\begingroup$ @mathe .And what has to do with the first question?Then check and see what happens. $\endgroup$ Jun 22, 2018 at 11:31

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