Find the limit following:

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


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!

  • $\begingroup$ Possibly related: math.stackexchange.com/questions/576110/… $\endgroup$ – abiessu Nov 26 '13 at 18:16
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    $\begingroup$ The numerical value is 1.2722249619362552835210450521628613228181075332403 , according to PARI. Using the inverse symbolic calculator, I found no closed expression. $\endgroup$ – Peter Nov 26 '13 at 18:28
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    $\begingroup$ n=10000;u=(1/n)^(1/n);while(n>2,n=n-1;u=(u+1/n)^(1/n));print(u) $\endgroup$ – Peter Nov 26 '13 at 18:51
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    $\begingroup$ 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... $\endgroup$ – Benjamin Dickman Nov 26 '13 at 20:06
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    $\begingroup$ 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! $\endgroup$ – Hamid Shafie Asl Jun 11 '14 at 11:20

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}$

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  • $\begingroup$ That's even better! I wonder if it is possible to get even tighter bound... $\endgroup$ – Aron D'souza Feb 20 '16 at 11:51
  • $\begingroup$ The next bound is approximately $L<1.272871$ $\endgroup$ – Yuriy S Feb 20 '16 at 11:54
  • $\begingroup$ Add it as an answer then. $\endgroup$ – Aron D'souza Feb 20 '16 at 11:55

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:


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


The bound will become:


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:

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

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


And solving the quintic equation:


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


Taking into account the corresponding lower boundary:


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:


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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 $\phi$.

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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.

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