How do you find the integer part of the following expresion?

Find the integer part of 1/sqrt1+1/sqrt2+1/sqrt3+....+1/sqrt1000000

1 Answer
Steve M
Apr 28, 2018

1999

Explanation:

We seek:

S = 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(1000000) =sum_(r=1)^1000000 \ 1/sqrt(r)

The sum:

S_n = sum_(r=1)^n \ 1/sqrt(r)

Is bounded from above by:

I_u(n) = int_0^n 1/sqrt(t) \ dt

and bounded from below by:

I_l(n) = int_1^n 1/sqrt(t) \ dt

Now:

int_a^b 1/sqrt(t) \ dt = [sqrt(t)/(1/2)]_a^b = 2(sqrt(b)-sqrt(a))

So we can write:

I_l(n) lt S_n lt I_u(n)

And with n=1000000, we have:

2(sqrt(1000000)-sqrt(1)) lt S lt 2(sqrt(1000000)-sqrt(0))

:. 2(sqrt(1000000)-1) lt S lt 2(sqrt(1000000))

:. 2sqrt(1000000)-2 lt S lt 2sqrt(1000000)

:. 2*1000-2 lt S lt 2*1000

:. 2000-2 lt S lt 2000

:. 1998 lt S lt 2000

Hence, the integer part of the sum, S, is 1999

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