# Two inequalities with $e$

I have problem because I can't do this exercise:

Prove that: $\forall x>1$, $(1+\frac{1}{x})^x<e<(1+\frac{1}{x-1})^x$

and: $\forall (p,q)>0$, $\left ( 1+\frac{p}{q} \right )^q\leq e^p \leq \left ( 1+\frac{p}{q} \right )^{p+q}$

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To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. – Zev Chonoles Jan 3 '13 at 18:02
Note that $e\lt (1-\frac{1}{x-1})^x$ doesn't look right when $x\gt 1$. – André Nicolas Jan 3 '13 at 18:05
Putting $x=2,e<(1-\frac1{2-1})^2=0??$ – lab bhattacharjee Jan 3 '13 at 18:05
If you take logarithms in the first inequality, replace $x$ by $1/x$, and multiply by $x$, you see that you have to prove $\ln(1+x)<x$. But $y=x$ is the tangent at $x=1$ of $y=\ln(1+x)$, and the latter curve is concave, so that is not too hard. A similar strategy may well work on the other inequalities. – Harald Hanche-Olsen Jan 3 '13 at 18:14
I think the second inequality could be $e<(1+1/x)^{x+1}$ then $\forall x>1$ can be changed to $\forall x>0$ as in the second inequality pair. This includes the original inequality and you can immediately see that the two inequality pairs are equivalent by taking the $p$-th root and subsituting $\frac{q}{p}$ by $x$ and changing the $\le$ by $\lt$. – miracle173 Jan 4 '13 at 12:39

The first inequality follows from the second with $p=1$, if we'll prove it with $<$ instead of $\le$.

Taking a $p$'th root of the second inequality, it becomes: $$(1+x)^{\frac{1}{x}} < e < (1+x)^{\frac{1}{x}+1}$$ Where $x=\frac{p}{q} >0$. Taking logarithm, it becomes $$\frac{1}{x}\ln(1+x) < 1 < (\frac{1}{x}+1)\ln(1+x)$$ or: $$\ln(1+x) < x< (1+x)\ln(x+1)$$

This can be proved by showing $x-\ln(1+x), (1+x)\ln(1+x)-x$ are increasing: $$(x-\ln(1+x))'=1-\frac{1}{x+1}=\frac{x}{x+1}>0 \implies$$ $$x-\ln(1+x) > 0 - \ln1 =0$$ $$((x+1)\ln(1+x)-x)'=\frac{x+1}{x+1}+\ln(x+1)-1 = \ln(x+1) > \ln 1 = 0 \implies$$ $$(x+1)\ln(1+x)-x > 1 \ln 1 - 0 = 0$$

As mentioned in the comments, this can be interpreted geometrically as follows: $\ln(1+x)$ is concave and $(1+x)\ln(1+x)$ is convex (with second derivative $\frac{-1}{(1+x)^2},\frac{1}{1+x}$ respectively), and $l(x) = x$ is tangent to both functions at $x=0$. The tangent to a convex graph is always below the graph, and for a concave graph it is always above. See this graph.

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Using the series $$-\log(1-u)=\sum_{k=1}^\infty\frac{u^k}{k}\tag{1}$$ we get that \begin{align} x\log\left(1+\frac1x\right) &=(1-(x+1))\log\left(1-\frac1{x+1}\right)\\ &=-\sum_{k=1}^\infty\frac1{k(x+1)^k}+\sum_{k=1}^\infty\frac1{k(x+1)^{k-1}}\\ &=-\sum_{k=1}^\infty\frac1{k(x+1)^k}+\sum_{k=0}^\infty\frac1{(k+1)(x+1)^k}\\ &=1-\sum_{k=1}^\infty\frac1{k(k+1)(x+1)^k}\\ &\lt1\quad\text{for }x\ge0\tag{2} \end{align} Similarly, \begin{align} x\log\left(1+\frac1{x-1}\right) &=-x\log\left(1-\frac1x\right)\\ &=\sum_{k=1}^\infty\frac1{kx^{k-1}}\\ &=1+\sum_{k=1}^\infty\frac1{(k+1)x^k}\\ &\gt1\quad\text{for }x\gt1\tag{3} \end{align} Combining $(2)$ and $(3)$, we get that for $x\gt1$, $$\left(1+\frac1x\right)^x\lt e\lt\left(1+\frac1{x-1}\right)^x\tag{4}$$

Let $x=\frac qp$, then $(2)$ says that for $p,q\gt0$, $$\left(1+\frac pq\right)^{q/p}\lt e\tag{5}$$ Let $x=1+\frac qp$, then $(3)$ says that for $p,q\gt0$, $$\left(1+\frac pq\right)^{(p+q)/p}\gt e\tag{6}$$ Combining $(5)$ and $(6)$ says that for $p,q\gt0$ $$\left(1+\frac pq\right)^q\lt e^p\lt\left(1+\frac pq\right)^{p+q}\tag{7}$$

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