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

The claim in your header is clearly false, as has been pointed out to you...so something is off. The body of your post speaks about factorials and has $n$ in an exponent, so it looks very different from the question in your header. — lulu, Jan 15 '24 at 16:07
$2n^6 < (n+1)^6 \iff 2 < \frac {(n+1)^6}{n^6}=(1+\frac 1n)^6\iff \sqrt[6] 2 < 1+\frac 1n\iff \sqrt[6]2-1 < \frac 1n\iff n < \frac 1{\sqrt[6]2 -1} \approx 8.165$ so for $n \ge 9$ this will not be true. [$1062882= 2\cdot 9^6 > 10^6= 1000000$ for example] — fleablood, Jan 15 '24 at 17:09
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — Henrik supports the community, Jan 15 '24 at 17:10
"on further solving at last i encountered the inequality 2(n⁶) < (n+1)⁶" How???? — fleablood, Jan 15 '24 at 17:30
This post (once corrected some more) will be a duplicate of https://math.stackexchange.com/questions/3353103 — Anne Bauval, Jan 16 '24 at 11:42

score 1 · Answer 1 · answered Jan 16 '24 at 03:14

1

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 $.

answered Jan 16 '24 at 03:14

T. Feix

121

A sad math boy · Accepted Answer · 2024-01-16T10:47:44.750

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.

Thank you very much if you can explain the derivative method it will be good — user4372, Jan 16 '24 at 10:05

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

2 Answers2