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Wlod AA
Wlod AA
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I stumbled upon the following inequality which, I believe, is true. I was able to prove it for small k, but I have no proof for the general case. Any help is welcome.

Let $n\geq k\geq 1$ then $$\left(1+\frac{1}{n}\right)^k S(n+1,k+1)\geq \left(1+\frac{1}{k}\right)^n S(n,k)$$$$\left(1+\frac{1}{n}\right)^k\cdot S(n+1,k+1)\,\ \geq \,\ \left(1+\frac{1}{k}\right)^n\cdot S(n,k)$$ where $S(n,k)$ is a Stirling number of second kind.

I stumbled upon the following inequality which, I believe, is true. I was able to prove it for small k, but I have no proof for the general case. Any help is welcome.

Let $n\geq k\geq 1$ then $$\left(1+\frac{1}{n}\right)^k S(n+1,k+1)\geq \left(1+\frac{1}{k}\right)^n S(n,k)$$ where $S(n,k)$ is a Stirling number of second kind.

I stumbled upon the following inequality which, I believe, is true. I was able to prove it for small k, but I have no proof for the general case. Any help is welcome.

Let $n\geq k\geq 1$ then $$\left(1+\frac{1}{n}\right)^k\cdot S(n+1,k+1)\,\ \geq \,\ \left(1+\frac{1}{k}\right)^n\cdot S(n,k)$$ where $S(n,k)$ is a Stirling number of second kind.

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Robert Z
Robert Z
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Inequality for Stirling numbers of the second kind

I stumbled upon the following inequality which, I believe, is true. I was able to prove it for small k, but I have no proof for the general case. Any help is welcome.

Let $n\geq k\geq 1$ then $$\left(1+\frac{1}{n}\right)^k S(n+1,k+1)\geq \left(1+\frac{1}{k}\right)^n S(n,k)$$ where $S(n,k)$ is a Stirling number of second kind.