Question 1183560
{{{lim(n->infinity, (1/n)*sum(k/sqrt(n^2+k^2), k=1,n))}}}

= {{{lim(n->infinity, (1/n)*sum((k/n)/(sqrt(1+(k/n)^2)), k=1,n))}}}


=  {{{lim(n->infinity, sum((1/n)*((k/n)/(sqrt(1+(k/n)^2))), k=1,n))}}}

The term inside the summation eerily seems like the area of a rectangle whose width is {{{1/n}}}, and whose height is {{{(k/n)/sqrt(1+(k/n))^2}}}.  
As such, this looks like the equipartition of the interval [0,1] in the construction of the upper Riemann sums for the function {{{f(x) = x/sqrt(1+x^2)}}}.  Hence,

 {{{lim(n->infinity, sum((1/n)*((k/n)/(sqrt(1+(k/n)^2))), k=1,n)) = int(x/sqrt(1+x^2), dx, 0,1)}}}.

The integral is easily evaluated as {{{sqrt(2)-1}}}.  Therefore, the infinite series converges and the sum is {{{sqrt(2) - 1}}}, approximately 0.41421 to 5 d.p.