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# lim n→∞  [x]+[2x]+[3x]+.........+[nx]                                   n2

12 years ago

I think the answer is                          X/2

heres how u do it :

[x] can be written as X+{x}    --> {x} is fractional part .. between 0 and 1

therefore ,

lim n→∞  [x]+[2x]+[3x]+.........+[nx]

n2

is =

lim n→∞  x+2x+3x+.........+nx + {x}+{2x} ...+{nx}

n2

= x(1+2+3+..+n)   +   {x}+...{nx}

n^2

=x(n(n+1))/2n^2 +  ({x}+...{nx})/n^2

Since {} is only b/w 0 and 1  , the second operand becomes 0 as n tends to

on solving the first part , u get (n2 x +nx)/2n2 = x/2 +x/2n

x/2n becomes 0 as x tends to infinity.

An easy way to do limits when Greatest integer  function is there, is to take out the commom element which does not change ... in this case x. (x , 2x, 3x ...)  x remains same.

divide this element by the power to which the denominator (having the dependent variable) is raised.

eg: [x]+[2x]+[3x]....+[nx]

n3

as n tends to infinity, = x/3