Question Number 21782 by drbl.gnbd last updated on 03/Oct/17
$${a}_{\mathrm{1}} =\mathrm{1},\:{a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} }{\:\sqrt{{a}_{{n}} +{n}+\mathrm{1}}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} =? \\ $$
a-1-1-a-n-1-a-n-a-n-n-1-n-1-a-n-