For a natural number k prove that 2(kᵏ) < (k+1)ᵏ

Question

For a natural number k prove that 2(kᵏ) < (k+1)ᵏ

I encountered this inequality while encountering an induction problem that stated 2ⁿ.n! < nⁿ For the induction step i chose 6 which satisfies Assume for some positive k>6
2ᵏ.k! < kᵏ ...(1)
holds true then we must show that

2ᵏ+¹.(k+1)! < (k+1)ᵏ+¹ follows from (1) multiplying eq(1) by 2.(k+1) on both side which results in 2ᵏ+¹.(k+1)! < 2kᵏ.(k+1)

If we can show that 2kᵏ< (k+1)ᵏ then it will complete the proof. CAN SOMEONE HELP

Answer 1

If I'm correctly understanding your problem,you are saying $2x^x<(x+1)^x$ when x > 1 and x a natural number.

Well if you expand the right hand side by the binomial theorem, you get $(x+1)^x=x^x+x\times x^{x-1}+C_x^2\times x^{x-2}+...$

Just taking the first two terms in the expantion gives you $(x+1)^x>2x^x$

Actually you can also prove this by derivatives

The original is equivalent to $2<(1+\frac{1}{x})^x$

Now let's prove that $((1+\frac{1}x)^x)'$ is bigger than zero when x is positive

It turns out that the above is equal to $e^{x\ln\left(1+\frac{1}{x}\right)}\left(-\frac{1}{x+1}+\ln\left(1+\frac{1}{x}\right)\right)$

The exponential part on the left is bigger than zero,so we just need to prove that $-\frac{1}{x+1}+\ln\left(1+\frac{1}{x}\right)>0$.This can be seen easily by taking another derivative.

Answer 2

I don't exactly understand what you are trying to say, but $2k^k < (k+1)^k; k \in \mathbb{N}$ can be disproven without induction by providing the counterexample $k = 1 \Rightarrow 2 \times1^1 = (1+1)^1 $.