Prove by induction that (n^2 + 4) / 4n is true for all n = 1, 2, 3, ....

I've proven the base case to be true, ie.
when n = 1
(1^2 + 4) / 4 = 5/4 = 1.25

Assume true for n = k, i.e.
(k^2 + 4) / 4k is true

When n = (k + 1)

((k + 1)^2 + 4) / (4k + 4)
however, I'm stuck right here. Can someone explain what I should do next. Thanks

Question

Prove by induction that (n^2 + 4) / 4n is true for all n = 1, 2, 3, ....

I've proven the base case to be true, ie.
when n = 1 
(1^2 + 4) / 4 = 5/4 = 1.25

Assume true for n = k, i.e.
(k^2 + 4) / 4k is true

When n = (k + 1)

((k + 1)^2 + 4)  / (4k + 4)
however, I'm stuck right here. Can someone explain what I should do next. Thanks

nnesha · Answer

i see there is a typo it should 1+4  over 4

anonymous · Answer

There was an error in my first post it should be (1^2 + 4) / 4

anonymous · Answer

The problem was a recurrence (0.5T(n/2) + 1/n). I had used repeated substitution to arrive at the equation above for the recurrence (k^2 + 4)/4k. Now I wanted to prove that the equation was indeed correct. Which takes you to the beginning of my question.

turingtest · Answer

What do you mean that "(n^2 + 4) / 4n is true?"
That is not an equation, so it can be neither true nor false.

anonymous · Answer

It works out to the value that of the recurrence, hence, T(1) = 1.25. I need to show that the equation holds for all values of n.

turingtest · Answer

(n^2 + 4) / 4n is not equal to 5/4 for n=2 or n=3, but it is for n=4
perhaps it is true for n=4k where k is in 1,2,3...  ?

anonymous · Answer

It won't equal 5/4 for those, that was only for T(1). For T(2) it should equal 1, and for T(2) = 13/12

turingtest · Answer

so you want to prove that (n^2 + 4) / 4n is equal to what exactly?