46
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Possibly related: math.stackexchange.com/questions/576110/… $\endgroup$ – abiessu Nov 26 '13 at 18:16
  • 7
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
3
$\begingroup$

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}$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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:

$$1.27207<L<1.27228$$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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 $\phi$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.