Prove that $\frac{(n+2)^{n+1}}{(n+1)^n}-\frac{(n+1)^n}{n^{n-1}}<e$

Question

For all $n\in\mathbb N$ prove that: $$\frac{(n+2)^{n+1}}{(n+1)^n}-\frac{(n+1)^n}{n^{n-1}}<e$$

We can rewrite it in the following form $$(n+2)\left(1+\frac{1}{n+1}\right)^{n}-(n+1)\left(1+\frac{1}{n}\right)^{n-1}<e$$ or $$(n+1)\left(1+\frac{1}{n+1}\right)^{n+1}-n\left(1+\frac{1}{n}\right)^{n}<e$$ and how should I proceed?

Answer 1

Claude Leibovici's answer prove only that the proposition is true for sufficiently large rather than arbitrary positive $n$. Here is the complete proof with stronger result.

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Let $$f(x):=x\left(1+\frac{1}{x}\right)^{x},\ x>0.$$ $$f'(x) = e^{x\ln(1+\frac1x)}\Big[1+x\Big(\ln\Big(1+\frac1x\Big)-\frac1{1+x}\Big)\Big].$$ $f$ increases since $$\ln\Big(1+\frac1x\Big)=-\ln\Big(1-\frac1{1+x}\Big)>\frac1{1+x}$$ by Taylor expansion of the logarithmic function around $1$. Because $f(x)$ is strictly convex, $1=f'(0)<f'(x)<f'(\infty)=e,\ \forall x\in (0,\infty)$.

So we conclude more than the required inequality $$1<f(n)-f(n-1)<f(n+1)-f(n) = \int_n^{n+1}f'(x)dx<e,\ \forall n\ge 1.$$

Answer 2

I do not know if this is a satisfactory answer for you.

Consider $$A=\frac{(n+2)^{n+1}}{(n+1)^n}\qquad , \qquad B=\frac{(n+1)^n}{n^{n-1}}$$ Take logarithms and expand as Taylor series for large values of $n$. This would lead to $$\log(A)=1+\log \left(n\right)+\frac{1}{2 n}+\frac{1}{3 n^2}-\frac{13}{12 n^3}+O\left(\frac{1}{n^4}\right)$$ $$\log(B)=1+\log \left(n\right)-\frac{1}{2 n}+\frac{1}{3 n^2}-\frac{1}{4 n^3}+O\left(\frac{1}{n^4}\right)$$ Taylor again $$A=e^{\log(A)}=e n+\frac{e}{2}+\frac{11 e}{24 n}-\frac{43 e}{48 n^2}+O\left(\frac{1}{n^3}\right)$$ $$B=e^{\log(B)}=e n-\frac{e}{2}+\frac{11 e}{24 n}-\frac{7 e}{16 n^2}+O\left(\frac{1}{n^3}\right)$$ $$A-B=e-\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right)$$

Answer 3

\begin{align*} f(n) &= (n+2)(1+\frac{1}{n+1})^n - (n+1)(1+\frac{1}{n})^{n-1}\\ &= (1+\frac{1}{n+1})^n + (n+1)((1+\frac{1}{n+1})^n - (1+\frac{1}{n})^{n-1}) \rightarrow e, \end{align*} If $a(n) = n(1+\frac{1}{n})^n$ then, because of convexity, $(a(n+1) - a(n-1))/2 > a(n)$ and $$ f(n+1) - f(n) = a(n+1) - 2a(n) - a(n-1) > 0. $$ Since $f(1)<e$ and the function is increasing it must approach the limit from below so $f(n) < e$.