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Let $a>0$ . How we can find the limit of :

$$\sqrt{a+\sqrt{2a+\sqrt{3a+\ldots}}}$$

Thanks in advance for your help

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    $\begingroup$ see en.wikipedia.org/wiki/Nested_radical (if only for a general introduction) $\endgroup$ – The Chaz 2.0 Jun 29 '11 at 19:40
  • $\begingroup$ And mathworld.wolfram.com/NestedRadical.html Formulas 26-28 have some resemblance $\endgroup$ – leonbloy Jun 29 '11 at 19:44
  • $\begingroup$ Define a sequence recursively. Then you can show the sequence is monotone increasing (since you assumed a>0), so that the limit exists. Then , by continuity of sqr, you can introduce the limit inside of the square root, and use the fact that $a_{n-1} \rightarrow {a_n}$ $\endgroup$ – gary Jun 29 '11 at 19:52
  • $\begingroup$ What do you mean by limit? Your expression seems to be between $\sqrt{a} +\sqrt{1/2}$ and $2a+1$, approaching the lower bound as $a$ increases and the higher bound as $a$ decreases towards $0$. $\endgroup$ – Henry Jun 29 '11 at 20:10
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    $\begingroup$ See Bill Dubuque's answer here: math.stackexchange.com/questions/7204/limit-of-nested-radical/… which mentions Vijayaraghavan's result. $\endgroup$ – Aryabhata Jun 29 '11 at 20:16
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In general, this is not known (as far as I know, at least). This is an extension of the so-called Kasner Number, and there is a good paper on the problems of finding closed forms of the Number can be found here (the paper also discusses infinite nested radicals).

This sequence does converge, at least. It's monotone increasing and bounded, so that's handy.

As an interesting note, many people are familiar with the work of Kasner without knowing it. It was Kasner who gave the name "googol" to the number 1 followed by 100 zeroes. Cool.

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    $\begingroup$ actually it was kasner's (9yrold at the time) nephew who gave $10^{100}$ its fancy name $\endgroup$ – yoyo Jun 29 '11 at 20:53
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    $\begingroup$ @yoyo: I heard! It's such a funny story. $\endgroup$ – davidlowryduda Jun 29 '11 at 21:00
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For the case $a=1$, see OEIS sequence A072449.

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Define $a_n$ to be the sequence presented. Then we have the following (for $n$ large enough such that $an>1$): $$ a_n < \sqrt{1+\sqrt{2+..+\sqrt{(n-1)a+na}}}\stackrel{not}=b_n$$

$$b_{n+1}=\sqrt{1+\sqrt{2+...+\sqrt{(n-1)a+\sqrt{na+(n+1)a}}}} $$

For $n$ large enough we have that $b_{n+1}\leq b_n$ since $na\geq \sqrt{(2n+1)a}$ for $n \geq N_0$. Since $a_n$ is bounded above by a decreasing sequence, and $(a_n)$ is clearly increasing, it is convergent. However, I think its limit does not have a closed form. Even the case $a=1$ presented in the other answer does not have a closed form.

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Define the sequence $a_n$ recursively by:

$a_1$:=$\sqrt{a}$

$a_n$= $\sqrt{a+a_{n-1}}$

This sequence is monotone non-decreasing, and bounded so the limit exists, and it is equal to y.

Now use the fact that $\sqrt{x}$ is continuous for $x>0$, so that it is sequentially-continuous;

then, lim

$\ \lim_{x\to\infty} \sqrt{a + \sqrt{a_{n-1}}} $=

$\sqrt{\ \lim_{x\to\infty}(a+\sqrt{a_{n-1}}) }$=

$\sqrt{a+\sqrt{\ \lim_{x\to\infty}(a_{n-1})}}$

By convergence, $a_{n-1}\rightarrow$y

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  • $\begingroup$ Wow, I think I finally got the formatting right!. I am so exhausted, I will nap for a whole week. $\endgroup$ – gary Jun 29 '11 at 20:24
  • $\begingroup$ You are defining a different sequence: $\sqrt{a + \sqrt{a + \dots}}$, isn't it? What you have defined has a closed form, as the root of a quadratic, I believe. See: math.stackexchange.com/questions/11945/… $\endgroup$ – Aryabhata Jun 29 '11 at 20:26
  • $\begingroup$ Yes, you are right, but I think a small change may save the amswer for this question $\endgroup$ – gary Jun 29 '11 at 20:53
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    $\begingroup$ How embarrassing to be so absent-minded and answer questions that haven't been asked. I'll switch from a single shotof espresson to a quadruppio. Sorry to all. $\endgroup$ – gary Jun 29 '11 at 20:56
  • $\begingroup$ Happens to all, don't worry :-) $\endgroup$ – Aryabhata Jun 29 '11 at 20:59

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