Why is $(1+k)^{\frac{|p|}{2}+1}\geq k(2+k)^{\frac{|p|}{2}}$ for $k\in\mathbb{N}$ and $-2<p<0$

Question

I am trying to prove that the sequence $b_k=\frac{1}{k(1+k)^{\frac{p}{2}}}$ is monotone decreasing for $k\in\mathbb{N}$ and $-2<p<0$. I am not sure how to do this. I was able to re-express the statement that $b_k$ is decreasing as the inequality: $$(1+k)^{\frac{|p|}{2}+1}\geq k(2+k)^{\frac{|p|}{2}}$$ But I cannot figure out how to prove that this inequality holds. Any help is appreciated.

Answer 1

The inequality $$(1+k)^{\frac{|p|}{2}+1}\geq k(2+k)^{\frac{|p|}{2}}$$ cannot hols for all $p$ with $|p| >2$. You can rewrite the inequality as $(1+\frac 1{k+1})^{|p|/2}\leq 1+\frac 1 k$. Letting $p \to -\infty$ leads to a contradiction.

For $-2<p<0$ the inequality $(1+\frac 1{k+1})^{|p|/2}\leq 1+\frac 1 k$ follow from the fact that $(1+\frac 1{k+1})^{|p|/2}\leq 1+\frac 1 {k+1} \leq1+\frac 1 k$.