<s>[INST] 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?
 [/INST] 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.
</s>