Question 29292: Show that 1/sqrt 1 + 1/sqrt 2 + 1/sqrt 3...+ 1/sqrt n < 2*sqrt n for all positive integers.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Show that 1/sqrt 1 + 1/sqrt 2 + 1/sqrt 3...+ 1/sqrt n < 2*sqrt n for all positive integers.
COSIDER THE EQN.
SQRT(N+1)-SQRT(N)={SQRT(N+1)-SQRT(N)}*{SQRT(N+1)+SQRT(N)/{SQRT(N+1)+SQRT(N)}
={(N+1)-(N)}/{SQRT(N+1)+SQRT(N)}=1 / {SQRT(N+1)+SQRT(N)}>1 / {SQRT(N+1)+SQRT(N+1)}
=1 / 2*SQRT(N+1)...HENCE
SQRT(N+1)-SQRT(N) > 1/2*SQRT(N+1)..PUT N=1,2,3...N..IN THIS EQN. AND ADD UP...
N=1...............SQRT2-SQRT1>1/2SQRT2
N=2...............SQRT3-SQRT2>1/2SQRT3
N=3...............SQRT4-SQRT3>1/2SQRT4
......................................
.......................................
N=N-1.............SQRT(N)-SQRT(N-1)>1/2SQRT(N)
N=N...............SQRT(N+1)-SQRT(N)>1/2SQRT(N+1)
-------------------------------------------------ADDING.....
WE FIND ALL TERMS ON LHS CANCEL EXCEPT
SQRT(N+1)-SQRT(1)>(1/2){ 1/sqrt 2 + 1/sqrt 3...+ 1/sqrt n }
OR....
{1/sqrt 1 + 1/sqrt 2 + 1/sqrt 3...+ 1/sqrt n } > 2*{SQRT(N+1)-SQRT(1)}+1/SQRT1
=2SQRT(N+1)-1>2SQRT(N+1)>2SQRT(N)
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