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Find the limit following:

$$L=\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots+\sqrt[\Large n]{\frac{1}{n}}}}$$

P.S

I tried to find the value of $\:L$, but I found myself stuck into the abyss of incertitude.

Thus, any help to get me out of this rift is more than welcome!

Yiorgos S. Smyrlis
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Iloveyou
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  • Possibly related: http://math.stackexchange.com/questions/576110/how-to-find-this-limit-a-lim-n-to-infty-sqrt1-sqrt-frac12-sqrt-fr?rq=1 – abiessu Nov 26 '13 at 18:16
  • @abiessu: I think it's not related to each other! – Iloveyou Nov 26 '13 at 18:25
  • That's okay, I didn't say that you asked a duplicate, I'm attempting to give an extra bit of information to your question so that others who find it might have other similar questions available to look at. – abiessu Nov 26 '13 at 18:28
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    The numerical value is 1.2722249619362552835210450521628613228181075332403 , according to PARI. Using the inverse symbolic calculator, I found no closed expression. – Peter Nov 26 '13 at 18:28
  • @Peter: How do you calculate that result? – Iloveyou Nov 26 '13 at 18:38
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    n=10000;u=(1/n)^(1/n);while(n>2,n=n-1;u=(u+1/n)^(1/n));print(u) – Peter Nov 26 '13 at 18:51
  • Starting with n=1000 gives the same result within this precision. – Peter Nov 26 '13 at 18:53
  • But this is only speculation, it's certainly not true! – Iloveyou Nov 26 '13 at 19:19
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    You could see if Landau's Algorithm (for denesting radicals) is of any help, but my guess is there isn't a "nice" expression for what you have here... – Benjamin Dickman Nov 26 '13 at 20:06
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    I guess its limit is 1. Because it is increasing sequence which is bounded(maybe) by a number less than 2. But I don't know how to prove it! – Hamid Shafie Asl Jun 11 '14 at 11:20
  • With the same backward recursion, implemented by a simple Pascal program, I've obtained exactly the same result as found by Peter. No doubt that it is correct. And it has no sense to duplicate this an answer of my own. – Han de Bruijn Jul 25 '14 at 20:37
  • Further analysis reveals that an upper bound for the error is better than $1/n!$ . See this answer for an idea how that works. – Han de Bruijn Jul 28 '14 at 19:46
  • Perhaps this infinite "expression" $(\ldots (((L^2-1)^2-\frac 12)^2-\frac 13)^2-\frac 14 \ldots )= 0$ is easier to handle? – flawr Jan 01 '16 at 10:43

4 Answers4

3

I like how Yiorgos S. Smyrlis approached to find upper limit of $L$. In similar way, you can easily observe $\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} > L$ and lets assume that it converges to some constant $c$. Now, we can write ,

$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} = c$

Squaring on both sides,

$\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\cdots}}} = c^2$

Which is nothing but,

$\frac{1}{2}+c = c^2$

Solving above quadratic expression, value of $c$ will be $\dfrac{1+\sqrt{3}}{2}$ . Thus we get slightly improved upper bound for $L$ as $ L < \dfrac{1+\sqrt{3}}{2}$

Community
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Aron D'souza
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Using Aron D'souza's idea further we can get:

$$L^2-\frac{1}{2}< \sqrt[3]{\frac{1}{3}+\sqrt[3]{\frac{1}{3}+\dots}}$$

$$\sqrt[3]{\frac{1}{3}+\sqrt[3]{\frac{1}{3}+\dots}}=\frac{1}{2} \left(1+\sqrt{\frac{7}{3}} \right)$$

$$L<\sqrt{1+\frac{1}{2} \sqrt{\frac{7}{3}}}=1.328067$$

To find the next bound we will need to solve:

$$c^4-c-\frac{1}{4}=0$$

The exact solution is too complex to write here (see Wolframalpha), so I'll just write it numerically:

$$\sqrt[4]{\frac{1}{4}+\sqrt[4]{\frac{1}{4}+\dots}}=1.0723501510383$$

The bound will become:

$$L<\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1.0723501510383}}=1.272871$$

Which is three correct digits of the numerical value of the limit.

To make my answer more complete, the exact value of $c_4$ is:

$$c=\frac{1}{2} \left( b+\sqrt{\frac{2}{b}-b^2} \right)$$

$$b=\sqrt{ \sqrt[3]{ \frac{a}{18} }-\sqrt[3]{ \frac{2}{3a} } }$$

$$a=9+\sqrt{93}$$

And solving the quintic equation:

$$c^5-c-\frac{1}{5}=0$$

We get the upper bound for the limit with four correct digits:

$$L<1.272282$$

Taking into account the corresponding lower boundary:

$$L>\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt[4]{\frac{1}{4}+\sqrt[5]{\frac{1}{5}+\sqrt[6]{\frac{1}{6}}}}}}=1.271035$$

We see that truncating the limit gives less accurate solutions than the method in this answer.

However, truncating at $\frac{1}{7}$ we can finally get very good boundaries:

$$1.27207<L<1.27228$$

Yuriy S
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This is a partial result:

The underlying sequence is increasing and upper bounded by $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}=\dfrac{1+\sqrt{5}}{2}=\phi$. Thus the limit exists and it is less than $\phi$.

Yiorgos S. Smyrlis
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0

Actually, there is another way which gives better upper boundary. I'm posting it as a separate answer because of the size.

First we notice that:

$$\lim_{n \to \infty} \left( \frac{1}{n} \right)^{\frac{1}{n}}=\lim_{n \to \infty} \left(1- \left(1- \frac{1}{n} \right) \right)^{\frac{1}{n}}=1$$

This is not a proof, but the fact is well known. Now let's consider the following:

$$1< \left(\frac{1}{n-1}+1 \right)^{\frac{1}{n-1}}<1+\frac{1}{(n-1)^2}$$

$$1< \left(\frac{1}{n-2}+1+\frac{1}{(n-1)^2} \right)^{\frac{1}{n-2}}<1+\frac{1}{(n-2)^2}+\frac{1}{(n-2)(n-1)^2}$$

On the next step we get:

$$\dots 1+\frac{1}{(n-3)^2}+\frac{1}{(n-3)(n-2)^2}+\frac{1}{(n-3)(n-2)(n-1)^2}$$

In the end we obtain the following inequality:

$$\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots}}<1+\sum^{\infty}_{k=2} \frac{1}{k ~k!}=Ei(1)-\gamma=1.3179$$

We can increase precision by moving the truncated series under the radical (and we should not forget to get rid of $2$ in every denominator):

$$1+2\sum^{\infty}_{k=3} \frac{1}{k ~k!}=1+2(Ei(1)-\gamma-1-1/4)=1.135804$$

$$L < \sqrt{\frac{1}{2}+1.135804}=1.27899$$

$$1+2\cdot 3 \sum^{\infty}_{k=4} \frac{1}{k ~k!}=1+6(Ei(1)-\gamma-1-1/4-1/18)=1.074080$$

$$L < \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1.074080}}=1.27305$$

Now for the lower boundary the better estimation would be:

$$L > \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+1}}=1.26517$$

This is not ideal, but much more accurate than just truncating the sequence.

Yuriy S
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