Sum of consecutive square roots inside a square root

Question

$$\large\sqrt{1+\sqrt{1+2+\sqrt{1+2+3+\sqrt{1+2+3+4+\cdots}}}}$$ 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?

Answer 1

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), $n = 1,2,3,...$

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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 $$\overline{\lim}\ {\left(\frac{\log{\frac{n(n+1)}{2}}}{2^n}\right)} = 0. $$

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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.)

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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}. $

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(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)

Answer 2

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.