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Prove $$\left(1+\frac{1}{\pi}\right)^{\pi+1}<\pi$$ without using calculator

I have tried to show that the derivative of $f(x)=x-\left(1+\frac{1}{x}\right)^{x+1}$ is greater than zero , at $x=\pi$ but, it is too hard for me.

Karl Kroningfeld
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cwk709394
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    Just a heads up. I attempted this using the mean value theorem for $g(x)=\ln(x^x)$. It failed to prove the inequality. – Karl Kroningfeld Nov 25 '13 at 12:31
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    Just to remind you, the derivative of a function f(x) at x=c is greater than zero, does not necessarily mean that f(c)>0, it means that the function is increasing at x=c. – Indrayudh Roy Nov 25 '13 at 12:44
  • It might make you afraid if you have that symbol $\pi$ there... (It happened for me).. Have you tried looking for some small values of $x$ if that equality holds? –  Nov 25 '13 at 12:56
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    @PraphullaKoushik Well, $\left(1+\frac1{x}\right)^{x+1}=x$ for $x \approx 3.14104152541079$ and WA gives us this plot. I'm not sure that the fear of $\Large{\pi}$ is the issue here. – Daniel R Nov 25 '13 at 13:03
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    @cwk709394 : Please explain what context this question came up in. I have thrown the kitchen sink at it, and I'm pretty certain there is no way to do this that does not involve a calculator. – Prahlad Vaidyanathan Nov 25 '13 at 13:08
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    Interesting problem. I haven't been able to solve it algebraically myself but here are my thoughts on this: Consider the two function $$y=x$$ and $$y=(1+\frac{1}{x})^{x+1}$$. Find their point of intersection wolfram shows this to be at a value of $x$ just below $\pi$. Then show that the second funtion is decreasing at this point. We already know that the first function is always increasing, therefore we know that the second function must be less than the first function at $x=\pi$. – Mufasa Nov 25 '13 at 13:11
  • Here is a visualisation of the point of intersection of these two curves and the line $x=\pi$. – Mufasa Nov 25 '13 at 13:18
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    @PraphullaKoushik : the question comes from the 131th formula http://mathworld.wolfram.com/PiFormulas.html – cwk709394 Nov 25 '13 at 13:18
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    The inequality is very sharp. We do need to perform calculations. If we assume that $3.1415 < \pi < 3.1416$ then we can show that $(1 + 1/\pi)^{\pi} < (1 + 1/3.1416)^{3.1416} = 2.38257\ldots$ and $\pi/(1 + 1/\pi) = \pi^{2}(\pi + 1) > 3.1415^{2}/(3.1416 + 1) = 3.3829\ldots$. The point I want to put is that it is sufficient to show that $3.1415 < \pi < 3.1416$ to establish this inequality. And conversely any $a \in (3.1415, 3.1416)$ satisfies $(1 + 1/a)^{a + 1} < a$. – Paramanand Singh Nov 26 '13 at 04:58

1 Answers1

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$$g(x)=\left(1+\frac{1}{x}\right)^{x+1}$$ is a decreasing function over $\mathbb{R}^+$ in virtue of the Bernoulli inequality.

Since $\frac{22}{7}>\pi$ (the Archimedean approximation), it is sufficient to show that $g(22/7)<\pi$.

Since $\left(1+\frac{7}{22}\right)^{29} < 3015$, it is sufficient to show that $\pi^7>3015$. This follows from $\pi>3.141$.

Jack D'Aurizio
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