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

lim n→∞  [x]+[2x]+[3x]+.........+[nx]




2 Answers

abhilash sainathan
14 Points
13 years ago

ans :x/2

anthony rebello
8 Points
13 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]


is =

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


= x(1+2+3+..+n)   +   {x}+...{nx}



=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.


therefore the answer is x/2.


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]


as n tends to infinity, = x/3




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