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#

Related questions
Impact of this question
1148 views around the world
You can reuse this answer
Creative Commons License