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0
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Hello, I have been stuck on this problem:

$$Q(n)=n /\sqrt{n}$$

Where n=(1,2,3,4.....)

DetermineProof all positive integers n  for which:      Q(n) > Q(n+1)

My thought:

By plugging in few numbers,1,2,3 etc

I noticed that Q(n)=sqrt(1),sqrt(2) etc..

Also that Q(1)= 1

Where in Q(n+1) if i plugged n=1

Q(2)=1

What i am trying to say but not sure how to write the proof or the idea mathematically: that Q(n+1) sequence starts at Q(2) so the sum of its sequences must be lower than Q(n) because Q(n) starts at Q(1)

For instance:

Q(1)=1  in q(n)

Q(2)=1 in q(n+1)

Sep 19, 2019

#1
0

Or did i do a mistake that Q(2)=1..

Q(2)=sqrt(2)

But the last statement is also correct in the case too

That Q(n+1) sequence starts at Q(2)

while Q(n) starts at Q(1)

so the sum of Q(n) terms must be > Q(n+1) terms..

Sep 19, 2019
#2
+104876
+2

n / √n  >   [ n + 1 ]  / √[n + 1]

n / √n   - [ n + 1 ] / √[n + 1]   > 0

n √[ n + 1] - [n + 1]√n

__________________  >  0

√n * √[n + 1]

n √[n + 1]  - [n + 1]√n  >  0

n √[n + 1]  > [ n + 1]√n        square both sides

n^2 [ n + 1]  > [n + 1]^2 * n

n^2 [ n + 1] - [n + 1]^2*n > 0

n [n + 1]  [ n - (n + 1 ) ]  > 0

n (n + 1] [ - 1] >  0

n ( n + 1) < 0

This is only true on   -1 < n < 0

So....no poistive  integers make this true

Also.....see the graph here : https://www.desmos.com/calculator/tmsohselsc

Q(n)  is never greater than Q(n + 1)

Sep 19, 2019