induction take2 (1 Viewer)

[...]

Prove that if n is a natural number then

1 + ( 1/sqrt(2) ) + ( 1/sqrt(3) ) + ... + ( 1/sqrt(n) ) >= sqrt(n)


>= means greater or equal to;

 

[shafqat]

Prove that if n is a natural number then

1 + ( 1/sqrt(2) ) + ( 1/sqrt(3) ) + ... + ( 1/sqrt(n) ) >= sqrt(n)


>= means greater or equal to;

Draw 1/sqrt(x+1). Draw vertical lines on each integer. Complete the rectangles so their areas are 1/sqrt2, 1/sqrt3, etc up to 1/sqrtn. Integrate from 0 to n-1.
Hence as the rectangles have a greater area than the integral,
( 1/sqrt(2) ) + ( 1/sqrt(3) ) + ... + ( 1/sqrt(n) ) > int 1/srt(x+1) from 0 to n-1 = 2sqrtn - 2
So 1 + ( 1/sqrt(2) ) + ( 1/sqrt(3) ) + ... + ( 1/sqrt(n) ) > 2sqrtn – 1 >= sqrtn for natural numbers. //

 

[nit]

Incidentally this is what we've just completed in maths so far this semester - riemann sums and the area as being the limit of the riemann sums.

 

[ngai]

Prove that if n is a natural number then

1 + ( 1/sqrt(2) ) + ( 1/sqrt(3) ) + ... + ( 1/sqrt(n) ) >= sqrt(n)


>= means greater or equal to;

topic says induction take2, so if u wanna use induction..
to get from n=k to n=k+1:
for n=k+1, u need 1 + ( 1/sqrt(2) ) + ( 1/sqrt(3) ) + ... + ( 1/sqrt(n) ) + 1/sqrt(n+1) >= sqrt(n+1)
which is true if u can prove sqrt(n) + 1/sqrt(n+1) >= sqrt(n+1), due to the assumption
and this is also true if u can prove sqrt(n(n+1)) + 1 >= n+1, ie. sqrt(n^2 + n) >= n
and this final thing is clearly true, since sqrt(n^2+n) >= sqrt(n^2) = n

write it up properly of course