SOLUTION: Does the series (1/√(n^2+1^2) + 2/√(n^2+2^2) +...+(n-1)/√(n^2+(n-1)^2)+ n/√(n^2+n^2))/n converge as n goes to infinity? If it does, what is the sum?

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Question 1183560: Does the series
(1/√(n^2+1^2) + 2/√(n^2+2^2) +...+(n-1)/√(n^2+(n-1)^2)+ n/√(n^2+n^2))/n
converge as n goes to infinity? If it does, what is the sum?
Found 2 solutions by ikleyn, robertb:
Answer by ikleyn(52790)   (Show Source): You can put this solution on YOUR website!
.
Does the series
(1/√(n^2+1^2) + 2/√(n^2+2^2) +...+(n-1)/√(n^2+(n-1)^2)+ n/√(n^2+n^2))/n
converge as n goes to infinity? If it does, what is the sum?
~~~~~~~~~~~~~~~~~~~


            This problem is not for beginning Calculus students.
            It is for mature/advanced Calculus students.
            Therefore, I will give my solution without going into details,  assuming that the reader has an adequate level. 


The sum is

     = .      (1)


Each term under the sum symbol can be estimated this way


     <=  =    from the top,                   (2)

and

     >=  =   from the bottom,       (3)

or  

     <=  <= .      (4)



It gives, in turn, the following estimations for the entire sums  


     <=  <= .      (5)


Next, each sum    contains the sum of arithmetic progression  1 + 2 + 3 + . . . + n = ,

therefore,    =  = .


From this point, estimations (5) can be re-written


     <=  <= .      (6)


Thus the terms    are asymptotically between    and  .



        The sequence    is, actually,  monotonically DECREASING  sequence.



From the first glance, it seems to be unexpected statement, but it is true and it can be strictly established formally.

Therefore, of the two estimations (6) from the top and from the bottom, the only BOTTOM estimation makes sense for us now.


So, we have the monotonically decreasing sequence    limited by the value    from the bottom --- THEREFORE,

     +--------------------------------------------+
     |   the sequence    is converged.         |
     +--------------------------------------------+



Answer by robertb(5830)   (Show Source): You can put this solution on YOUR website!

=

=
The term inside the summation eerily seems like the area of a rectangle whose width is , and whose height is .
As such, this looks like the equipartition of the interval [0,1] in the construction of the upper Riemann sums for the function . Hence,
.
The integral is easily evaluated as . Therefore, the infinite series converges and the sum is , approximately 0.41421 to 5 d.p.


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