Limit of an infinite summation The above is from Maple 2019.1. Is there a way to achieve the same result from MMA12?

Tried

Limit[Sum[Sqrt[1 + k^2/n^3] - 1, {k, 1, n}], {n -> Infinity}]

which did not work. Best I can get is a numerical approximation.

With[{n = 10^200}, NSum[Sqrt[1 + k^2/n^3] - 1, {k, 1, n}]]

0.166667

which is informative.

• Can try this With[{n = 10^200}, NSum[Sqrt[1 + k^2/n^3] - 1, {k, 1, n}]] // Rationalize Aug 22 '19 at 1:02
• @PlatoManiac Nice cheat :) Aug 22 '19 at 1:05
• I totally agree ;) Aug 22 '19 at 1:11
• With[{n = 10^220}, NSum[Sqrt[1 + k^2/n^3] - 1, {k, 1, n}]] yields -0.333333 in Mathematica 12.0.
– mjw
Aug 22 '19 at 1:27
• @mjw - For v12 use With[{n = 10^220}, Block[{\$MaxExtraPrecision = 700}, NSum[Sqrt[1 + k^2/n^3] - 1, {k, 1, n}, WorkingPrecision -> 15]]] // Rationalize Aug 22 '19 at 4:09

Some trickery: introduce a regulator to improve the convergence, which allows you to expand in $$1/n$$, and finally take the limit explicitly:

Series[(Sqrt[1 + k^2/n^3] - 1) E^(-ϵ k), {n, ∞, 6}]

$$\left(\sqrt{\frac{k^2}{n^3}+1}-1\right) e^{-\epsilon k}=e^{-\epsilon k}\left(\frac{k^2}{2 n^3}-\frac{k^4 }{8 n^6}+\mathcal O(n^{-7})\right)$$

Sum[(E^(-k ϵ) k^2)/(2 n^3) - (E^(-k ϵ) k^4)/(8 n^6), {k, 1, n}]
Limit[%, ϵ -> 0]

$$\sum_{k=1}^n\left(\sqrt{\frac{k^2}{n^3}+1}-1\right) e^{-\epsilon k}=\frac{40 n^5+54 n^4+5 n^3-10 n^2+1}{240 n^5}+\mathcal O(\epsilon)$$

Limit[%, n -> ∞]

$$\color{red}{\frac{1}{6}}$$

It's a shame that AsymptoticSum doesn't work out of the box for this problem. We can help it along by expanding the expression around $$n=\infty$$ and then using AsymptoticSum. Here is the expansion:

coeff = SeriesCoefficient[Sqrt[1 + k^2/n^3] - 1, {n, Infinity, m}];
coeff //TeXForm

$$\begin{cases} \binom{\frac{1}{2}}{\frac{m}{3}} k^{2 m/3} & (m \bmod 3)=0\land m>0 \\ 0 & \text{True} \end{cases}$$

The only nonzero terms are when $$m$$ is a positive multiple of 3. Let's check by resumming the individual terms:

Sum[k^(2 m)/n^(3 m) Binomial[1/2, m], {m, Infinity}]

-1 + Sqrt[(k^2 + n^3)/n^3]

Next, let's sum up the general coefficient using AsymptoticSum to leading order:

terms[m_] = Assuming[
m ∈ Integers && m > 0,
Apart[
FullSimplify @ AsymptoticSum[k^(2 m)/n^(3 m) Binomial[1/2, m], {k, 1, n}, {n, Infinity, 1}],
n
]
];
terms[m] //TeXForm

$$\frac{\binom{\frac{1}{2}}{m} n^{1-m}}{2 m+1}+\frac{1}{2} \binom{\frac{1}{2}}{m} n^{-m}$$

Clearly, the leading term is order $$n^{1-m}$$ and comes from the first coefficient:

terms

1/6 + 1/(4 n)

The $$1/n$$ dependence is incomplete as we need to include the $$m=2$$ coefficient as well.

terms

-1/(16 n^2) - 1/(40 n)

To get the $$n$$ series to a particular order, we can series expand the expression, and then use AsymptoticSum on the series expansion to the right order:

expansion[order_] :=
AsymptoticSum[
Normal @ Series[Sqrt[1 + k^2/n^3] - 1, {n, Infinity, order}],
{k, 1, n},
{n, Infinity, Max[Floor[order/3] - 1, 1]}
]

For example:

expansion //TeXForm

$$-\frac{651}{66560 n^5}+\frac{5}{352 n^4}-\frac{17}{1152 n^3}+\frac{5}{168 n^2}+\frac{9}{40 n}+\frac{1}{6}$$