IIT JEE , AIEEE Forum - Mathematics , Algebra

ramkumar_november

if

X = 1 + 1/

2 + 1/

3+ 1/

4+ 1/

5+ ..................... + 1/

10⁶.

find [X] where [ ] : denotes greatest integer function

ramkumar_november

atleast someone try it!!!!!!!

dip_xaverian

is the answer 1?

ramkumar_november

no the answer is 1998

konichiwa2x

very nice problem!

k..

(n=1000) Here is the proof.

let

consider this function in the interval

For every

, we have a rectangle with the interval

as its base and

as its height. The sum of the areas of these (upper) rectangles is certainly bigger than the area under the graph of

Now:
Sum of the areas of the rectangles =

, and
area under the graph of f(x) =

So we have

this time consider the rectangles with, again, the intervals

as their bases but this time

as their heights. this time the sum of the areas of these (lower) rectangles is certainly less than the area under the graph of

Now,
Sum of the areas of the rectangles =

Thus

So we've proved that

=

konichiwa2x

(here is the general formula for those who asked)

exactly the same argument shows that in general for any

we have

note that if

is not a perfect square, then in general these bounds are not good enough to give us

ramkumar_november

exact proof konichiwa ...................
this sum was given by my iit maths professor............
i was astounded when he gave the proof.
so i just wanted to share this problem with my goiitan friends...........

another simple method to solve this problem is........

the general formula is

2*(

(n+1) -

n ) < 1/

n < 2*(

n -

(n-1) )

applying summation from 1 to 10⁶ we get the answer........

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