# Sum of consecutive square roots inside a square root

$$\large\sqrt{1+\sqrt{1+2+\sqrt{1+2+3+\sqrt{1+2+3+4+\cdots}}}}$$ I saw this somewhere in the internet but, the website didn't provide me any further information. What is the sum of the equation above? What is it called?

• This kind of thing is called a "nested radical" or a "continued radical", and searching for those terms may turn up this example. Be sure to report back if you find anything. – Gerry Myerson Jan 18 '14 at 3:37
• It is approximately equal to 1.864 458 958 163 488 132 352 003 715 – Antonio Vargas Jan 18 '14 at 3:51
• @AntonioVargas. May I ask how you arrived to this result ? I am more conerned by the procedure than by the "approximate" result. Thanks. – Claude Leibovici Jan 18 '14 at 8:20
• @ClaudeLeibovici, certainly. I asked Mathematica to compute the result to 35 digits of accuracy using this code, which starts from the innermost element, $\sqrt{m(m+1)/2}$, and works outward through the radicals. To get the 28 digits I wrote in my comment one needs to calculate at least the 24th element in the sequence (using this method). I have no error bounds, though, and the only evidence I have that they're correct is that if I increase $m$ by 1 then those digits remain unchanged. – Antonio Vargas Jan 18 '14 at 16:40
• @AntonioVargas. It is just incredible to see what we can do with 0's and 1's ! Thanks for the code; it is beautiful. Cheers. – Claude Leibovici Jan 18 '14 at 16:43

What is the sum of the equation above? What is it called?

It's neither an equation nor a sum, but is a continued radical defined as the limit (as $n\to\infty$) of the sequence of the nested radicals $$u_n = \sqrt{a_1+\sqrt{a_2 + \cdots + \sqrt{a_n}}}$$ where $a_n = \sum_{i=1}^n i = n(n+1)/2$ (the $n^{th}$ triangle number), $n = 1,2,3,...$

That this limit exists follows from a theorem proved by T. Vijayaraghavan (1927):

For any sequence of nonnegative reals $(a_n)_{n=1,2,3,...}$, the sequence of nested radicals $(u_n)_{n = 1,2,3,...}$ with $$u_n = \sqrt{a_1+\sqrt{a_2 + \cdots + \sqrt{a_n}}}$$ converges if and only if there exists a finite upper limit $$\overline{\lim}\ {\left(\frac{\log{a_n}}{2^n}\right)} < \infty.$$

This clearly holds in the present case, because $$\overline{\lim}\ {\left(\frac{\log{\frac{n(n+1)}{2}}}{2^n}\right)} = 0.$$

To bound the error of $u_n$ as an approximation of the limit, the following consequence of a theorem proved by Herschfeld (1935) can be used:

If $a_i \gt 0 \ \ (i = 1,2,3,...)$, then for all $n \ge 1$, $$0 \le u_{n+1}-u_n \le \frac{1}{2^n}\sqrt{\frac{a_{n+1}}{a_1 a_2 \cdots a_n}}.$$

Substituting $a_i = i(i+1)/2$ and simplifying then gives $$0 \le u_{n+1}-u_n \le \frac{1}{n!}\sqrt{\frac{n+2}{2^{n+1}}} ,$$ whence, noting that $n! \ge \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$, $$0 \le u_{n+1}-u_n \le A \left(\frac{B}{n}\right)^n$$ where $A = \sqrt{\frac{3}{4\pi}} = 0.4886...$ and $B = \frac{e}{\sqrt{2}} = 1.922...$

Now for all $m > n > 1$,

\begin{align} 0 \le u_m - u_n & = (u_m - u_{m-1}) + (u_{m-1} - u_{m-2}) + \cdots + (u_{n+1} - u_n) \\ & \le A \left ( \left (\frac{B}{m-1}\right )^{m-1} + \left (\frac{B}{m-2}\right )^{m-2} + \cdots + \left (\frac{B}{n}\right )^{n} \right )\\ & \le A \left ( \left (\frac{B}{n}\right )^{m-1} + \left (\frac{B}{n}\right )^{m-2} + \cdots + \left (\frac{B}{n}\right )^{n} \right )\\ & \le A \left (\frac{B}{n}\right )^n \frac{1 - (\frac{B}{n})^{m-n}}{1 - \frac{B}{n}}\tag{*} \end{align} where the geometric progression was summed in the last step (noting that $B/n < 1$ for all $n > 1$).

Finally, taking limits with $m \to \infty$ and letting $u = \lim_{m \to\infty}{u_m}$, $$0 \le u - u_n \le A \left (\frac{B}{n}\right )^n \frac{1}{1 - \frac{B}{n}}.$$

(Note that (*) above also gives another proof of the existence of the limit, by showing the sequence $(u_n)_{n = 1,2,3,...}$ to be Cauchy.)

The latter error bound is rather weak, but it suffices to prove, for example, that

$$u = 1.8644589581634881323520037152739437841564220698266...$$

by computing $u_n$ for sufficiently large $n$. For example, $u_n$ has at least $n$ correct digits for any $n \ge 20$, because in that case $A \left (\frac{B}{n}\right )^n \frac{1}{1 - \frac{B}{n}} \le 0.5 \ 10^{-n}.$

(Aside)

Here's the Sage program I used to compute the $u_n$:

def u(n):
r = 0
for k in [1..n][::-1]: r = sqrt(k*(k+1)/2 + r)
return r

for n in [10..100]: print n, u(n).n(digits=100)


As Gerry Myerson said this type of problems are called nested radical.

Let us try the following method: Writing the sequence of partials $$D_n:= \sqrt {1 + \sqrt {1 + 2 + \sqrt {1 + 2 + 3 + \sqrt {1 + 2 + 3 + 4 + \ldots + \sqrt { \ldots + \sqrt { \ldots + \sqrt {1 + 2 + 3 + \ldots + n} } } } } } }$$

Labeling the terms of the series beginning from the last one up to the first one; denoted by $y_1^2,y_1^2,\cdots,y_{n-1}^2,y_n^2$, i.e., \begin{align} &y_n^{2}=1 + 2 + 3 + \ldots + n = \sum\limits_{k = 1}^n {k}=\frac{n(n+1)}{2} \\ &y_{n-1}^{2}=1 + 2 + 3 + \ldots + (n-1) = \sum\limits_{k = 1}^{n-1} {k}=\frac{n(n-1)}{2} \\ &y_{n-2}^{2}=1 + 2 + 3 + \ldots + (n-2) = \sum\limits_{k = 1}^{n-2} {k}=\frac{(n-1)(n-2)}{2} \\ &\vdots \\ &y_{3}^{2}=1 + 2 + 3 = \sum\limits_{k = 1}^{3} {k}=\frac{(3)(4)}{2}=6 \\ &y_{2}^{2}=1 + 2 = \sum\limits_{k = 1}^{2} {k}=\frac{(2)(3)}{2}=3 \\ &y_{1}^{2}=1 = \sum\limits_{k = 1}^{1} {k}=\frac{(1)(2)}{2}=1 \end{align} Thus, \begin{align} D_n = \sqrt {y_1 + \sqrt {y_2 + \sqrt {y_3 + \ldots + \sqrt { \ldots + \sqrt { \ldots + \sqrt {y_{n - 2} + \sqrt {y_{n - 1} + \sqrt {y_n } } } } } } } } \end{align} Now, let us divide $D_n$ into the following subsequences \begin{align} & S_{n,1}^2 = y_n \\ & S_{n,2}^2 = y_{n - 1} + S_{n,1} \\ &S_{n,3}^2 = y_{n - 2} + S_{n,2} \\ &S_{n,4}^2 = y_{n - 3} + S_{n,3} \\ &\vdots \\ &S_{n,n - 2}^2 = y_2 + S_{n,n - 1} \\ &S_{n,n - 1}^2 = y_1 + S_{n,n} \end{align} Finally, we have \begin{align} D_n &= S_{n,1}^2 + S_{n,2}^2 + S_{n,3}^2 + S_{n,4}^2 + \ldots + S_{n,n - 2}^2 + S_{n,n - 1}^2 \\ &= \sum\limits_{j = 1}^{n - 1} {S_{n,j}^2 } \\ &= \sum\limits_{j = 1}^n {y_j } + \sum\limits_{j = 1}^n {S_{n,j} } \end{align} I think this enough to give an indication.

• Sorry, where is the relation to the question? – flonk Jul 1 '14 at 10:29
• Thanks, i made a correction – Mohammad W. Alomari Jul 1 '14 at 12:16
• It's still not clear to me, how you "divide $D_n$" into the $S^2_{n,j}$, the radicals just disappeared. Also, I think your result is wrong as it has $\lim_{n\to\infty}D_n=\infty$ but we know that a finite limit should exist $\approx 1.8644....$. – flonk Jul 1 '14 at 13:45
• I quote here, "in 1966. Vijayaraghavan proved that a sufficient criterion for the convergence of the following sequence \begin{align} \sqrt {a_1 + \sqrt {a_2 + \ldots + \sqrt {a_n } } } \end{align} s that $\overline {\mathop {\lim }\limits_{n \to \infty } } \frac{{\log a_n }}{{2^n }} < \infty$. In our problem, $a_n=y_n=\frac{n(n+1)}{2}$, so that $\overline {\mathop {\lim }\limits_{n \to \infty } } \frac{{\log a_n }}{{2^n }} = \overline {\mathop {\lim }\limits_{n \to \infty } } \frac{{\log {\textstyle{{n\left( {n + 1} \right)} \over 2}}}}{{2^n }} = 0$ – Mohammad W. Alomari Jul 2 '14 at 8:48
• Oohh, i quote from (math.stackexchange.com/questions/7204/…) see also (math.stackexchange.com/questions/61048/…), nice discussion. – Mohammad W. Alomari Jul 2 '14 at 8:51