Question

Proof by mathematical induction that
(n + 1)^n < n^(n+1) for any integer n >= 3.

So i've checked the base case, 64<81 but i'm not finding a way to show that (k + 1)^k < k^(k+1) implies (k + 2)^(k+1) < (k+1)^(k+2)

Thanks.

Answer 1

you trying show it by graphing?

Answer 2

Im supposed to do it by induction, not graphing.

Answer 3

So you know to assume
\[(k+1)^k < k^{k+1}\]

we want to show \[(k+2)^{k+1}<(k+1)^{k+2}\]

So I guess we can try to multiply our assumption on both sides by first k+1 and see what we have then

Answer 4

You might also find it useful after that to realize k<k+1

Answer 5

@Juarismi
are you still there?

Answer 6

So\[(k+1)^{k+1}<k ^{k+2}+k ^{k+1}\]
given k<k+1 then (k+1)^(k+1)>k^(k+1)

How that help?

Answer 7

Yeah I was missing the two...
This problem might be a little harder than I though.

You might have to rewrite the problem:
Here I will help help you rewrite it:
So we are supposing \[ (k+1)^k<k^{k+1} \]
Now we are going to rewrite this true statement so whatever we rewrite it as we still get to assume it holds (because it is just a rewrite after all )

And I'm rewriting it because the multiplying of (k+1) on both sides wasn't very helpful.

So I notice there is powers on both sides. I'm going to try to rewrite it so the power k is only on one side

so we have
\[(k+1)^k < k^k k => (\frac{k+1}{k})^k <k => (1+\frac{1}{k})^k < k \]
Now you want to show
\[\text{ if } (1+\frac{1}{k})^k <k \text{ then } (1+\frac{1}{k+1})^{k+1}<k+1 \]

This should help

Answer 8

Try to use our assumption
notice in our assumption we have 1/k
and then thing we want to show as 1/(k+1)
but you know 1/k>1/(k+1)
because 1/(k+1) gets closer to zero a lot quicker than 1/k as k goes to infinity