# possible bug with Limit

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.

• What is the value of i? Jun 22, 2018 at 7:37
• Yes, possible a bug. It should be returns unevaluated. Jun 22, 2018 at 8:52
• 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. Jun 22, 2018 at 9:38
• For an approximation: Limit[Sum[ Series[Sqrt[1 + i^2/n^2]/n, {n, Infinity, 125}] // Normal, {i, n}], n -> Infinity] // N Jun 22, 2018 at 21:37
• Now returns unevaluated in 13.3.0. Sep 2, 2023 at 5:00

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 *)

• 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])`. Jun 22, 2018 at 10:30
• @mathe .And what has to do with the first question?Then check and see what happens. Jun 22, 2018 at 11:31