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?

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.
