- #1

murshid_islam

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- TL;DR Summary
- My proof by induction looks way too contrived. Is there a way to simplify it?

I'm trying to prove the statement ##n^2 + 1 < n!## for ##n \geq 4##. My proof by induction looks way too contrived. Is there a way to simplify it? Here's what I got.

For n = 4, ##n^2 + 1 = 17 < 4!##. So, the statement is true for n = 4. Now let's assume it's true for n = k, that is, ##k^2 + 1 < k!##

Now, for n = k+1,

##(k+1)^2 + 1 ##

##= (k^2 + 1) + 2k + 1##

##< k! + 2k + 1## ; [from the induction hypothesis for n = k]

##< k\cdot k! + 2k + 1##

##= (k+1-1)k! + 2k + 1##

##= (k+1)k! - k! + 2k + 1##

##= (k+1)! + 2k + (1 - k!)##

##< (k+1)! + 2k - k^2## ; [because 1 – k! < –k

##< (k+1)! + 0## ; [2k – k

Therefore, we have ##(k+1)^2 + 1 < (k+1)! ##

I was wondering if there's an easier way to do the n = k + 1 part.

For n = 4, ##n^2 + 1 = 17 < 4!##. So, the statement is true for n = 4. Now let's assume it's true for n = k, that is, ##k^2 + 1 < k!##

Now, for n = k+1,

##(k+1)^2 + 1 ##

##= (k^2 + 1) + 2k + 1##

##< k! + 2k + 1## ; [from the induction hypothesis for n = k]

##< k\cdot k! + 2k + 1##

##= (k+1-1)k! + 2k + 1##

##= (k+1)k! - k! + 2k + 1##

##= (k+1)! + 2k + (1 - k!)##

##< (k+1)! + 2k - k^2## ; [because 1 – k! < –k

^{2}from the induction hypothesis for n = k]##< (k+1)! + 0## ; [2k – k

^{2}= k(2 – k) < 0 for k > 2]Therefore, we have ##(k+1)^2 + 1 < (k+1)! ##

I was wondering if there's an easier way to do the n = k + 1 part.

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