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?


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)

