1Arshad seems correct..
prove 3^{n}>n^{3} for all positive integers `n`.
note: open for all
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7 Answers
13^n will alwayz be postive so....it will alwayz be greater in the 2nd and third quadrants....and for the 1st quadrant 3n will alwayz increase more than than n^3 ....an no intersection points.......so this inequality holds......
11I did not get how u guys plug in 2nd and 3rd quadrants......n is a natural number!!!
1so then it implies that this inequality holds for all real n
so it will also hold for all positive integers n.....
am i correct sir???
62yes arshad.. if somehting holds for all reals, it will hold for integers..
if something holds for all +ve numbers, it will hold for natural numbers too
But you have to prove that too
1Consider f(x) = x1/x
f '(x) = x1/x(1 - lnx)x2
for x>3 : f '(x) < 0
so f(x) is decreasing for x>3
for x>3 : f(x) < f(3)
=> x1/x < 31/3
=> x3 < 3x
It is proved for x>3
for x=1,2,3 we can directly check by putting values
so the inequality is true for all positive integers
