6

lim

Limit[Sum[2(2k)^(1/(2k))-k^(1/k),{k,n+1,2n}]-n, n -> ∞]

to solve by hand, $$\sqrt[y]y=e^{\frac{\ln y}{y}}\sim1+\frac{\ln y}{y}$$ $$2\sqrt[2k]{2k}-\sqrt[k]k\sim1+\frac{\ln2}{k}$$ $$\sum_{k=n+1}^{2n}\left(2\sqrt[2k]{2k}-\sqrt[k]k\right)\sim n+\ln2\sum_{k=n+1}^{2n}\frac1k\sim n+\ln 2 \, \int_{n+1}^{2n}\dfrac{1}{t}dt$$ $$\lim_{n\to \infty}\left(\sum_{k=n+1}^{2n}\left(2\sqrt[2k]{2k}-\sqrt[k]k\right)-n\right)=\ln^22$$

y-ung'1 0'm-o
  • 51

2 Answers2

9

As I wrote, this is rather math than Mathematica. Let me improve the Bob Hanlon's approach. First,

Series[2 (2 k)^(1/(2 k)) - k^(1/k), {k, Infinity, 1}]

$ O\left(\left(\frac{1}{k}\right)^2\right)+\frac{\log (2)}{k}+1$

Second, the sum of $O(k^{-2})$ over $k$ from $n+1$ to $2n$ is $O(n^{-1})$ so the one tends to zero as $n$ approaches $\infty$. We need estimates to ground it and this is math. Now

Sum[Normal[Series[2 (2 k)^(1/(2 k)) - k^(1/k), {k, Infinity, 1}]], {k,
n + 1, 2*n}] - n

$ \log (2) \psi ^{(0)}(2 n+1)-\log (2) \psi ^{(0)}(n+1)$

The last step is

Limit[%, n -> Infinity]

$\log ^2(2)$

To be sure,

N[%]

$ 0.480453$

user64494
  • 5,811
  • @ user13892: Thank you for your personal opinion. I don't think thatAsymptoticSum[ 2 (2 k)^(1/(2 k)) - k^(1/k), {k,n+1,2n}, {n,[Infinity],1} ]-n should have produced -(Log[2]/(4 n)) + Log[2]^2 . BTW, what do you mean by SW which is not a standard abbreviation? – user64494 Jan 27 '20 at 20:10
  • Typo in my second comment: Series not Sum. – user13892 Jan 27 '20 at 20:47
  • Isn't AsymptoticSum a behaviour of summed expression around a point approximated to a certain degree like Series? So why wouldn't it had produced that result? – user13892 Jan 27 '20 at 20:24
  • SW = Stephen Wolfram. I distinctly remember that in one of his live stream, he argued with his developers to have AsymptoticSum return SeriesData rather than the actual expression to have consistency with Series. Or maybe I am confusion it with something else. – user13892 Jan 27 '20 at 20:17
  • I think AsymptoticSum[ 2 (2 k)^(1/(2 k)) - k^(1/k), {k,n+1,2n}, {n,\[Infinity],1} ]-n should have produced -(Log[2]/(4 n)) + Log[2]^2 or more completely -(Log[2]/(4 n)) + Log[2]^2+O[1/n]^2 (there was fight between SW and others to have the output of Sum and AsymptoticSum be SeriesData in one of the liveCEOs stream). I think it is a bug or since it is experimental they haven't put in all the algorithms. The entire purpose of AsymptoticSum is defeated if you have to explicitly give the asymptotic behavior of summand. – user13892 Jan 27 '20 at 20:00
  • The command AsymptoticSum[ Normal[Series[2 (2 k)^(1/(2 k)) - k^(1/k), {k, Infinity, 1}]], {k, n + 1, 2n}, {n, Infinity, 2}] - n performs $\frac{\log (2)}{16 n^2}-\frac{\log (2)}{4 n}+\log ^2(2) $. We need take the limit and this is longer. The command AsymptoticSum[ 2 (2 k)^(1/(2 k)) - k^(1/k), {k, n + 1, 2n}, {n, Infinity, 2}] - n fails. Don't hesitate to ask for further explanation in need. – user64494 Jan 27 '20 at 19:09
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    Wonderful! Do you know what is the purpose of AsymptoticSum? Does it deal with this kind of situation? – user13892 Jan 27 '20 at 19:02
4

It looks like a very complicated sum but it does seem to slow done enough to estimate its value and may converge!

Mathematica is not able to solve it analytically in my attempt.

First, a few things to note:

  • n-th root of x in real domain is represented by Surd in Mathematica.
  • Secondly, you need to take the discrete limit over the integers here.

Now you can try the analytical approach as follows:

ClearAll[f];
f[n_]:=Sum[2Surd[2k,2k]-Surd[k,k],{k,n+1,2n}]-n;

DiscreteLimit[
    f[n],
    n->∞
]

Maybe someone here can help with how to make Mathematica get to the analytical limit.

But we can estimate its value:

list=Table[
    N@f[n],
    {n,1,1000}
];

ListPlot[list,PlotRange->All]
ListPlot[Differences@list,PlotRange->All]

Estimation:

N@f[10000]
N@f[100000]
N@f[1000000]

0.479474

0.480298

0.48043

user13892
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