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)
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..
help please asap
+128406 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)
