An upper bound for a series

• May 15th 2009, 02:42 AM
NonCommAlg
An upper bound for a series
This question reminded me of a better one: Prove that $\sum_{n=1}^{\infty} \left(1 - \frac{1}{\sqrt{n}} \right)^n < \frac{1}{\sqrt{e}} \zeta \left(\frac{e}{2} \right).$

Spoiler:
Try to prove this stronger result: for all real numbers $x \geq 1: \ \left(1 - \frac{1}{\sqrt{x}} \right)^x < \frac{1}{\sqrt{ex^e}}.$
• Jun 1st 2010, 09:51 PM
chiph588@
Quote:

Originally Posted by NonCommAlg
This question reminded me of a better one: Prove that $\sum_{n=1}^{\infty} \left(1 - \frac{1}{\sqrt{n}} \right)^n < \frac{1}{\sqrt{e}} \zeta \left(\frac{e}{2} \right).$

Spoiler:
Try to prove this stronger result: for all real numbers $x \geq 1: \ \left(1 - \frac{1}{\sqrt{x}} \right)^x < \frac{1}{\sqrt{ex^e}}.$

I'll prove what's in the spoiler box (the stronger result):

$\left(1 - \frac{1}{\sqrt{x}} \right)^x < \frac{1}{\sqrt{ex^e}} \iff \log\left(\sqrt{ex^e}\left(1 - \frac{1}{\sqrt{x}} \right)^x\right) < 0$

$\log\left(\sqrt{ex^e}\left(1 - \frac{1}{\sqrt{x}} \right)^x\right)=\frac12+\frac12e\log(x)+x\log\lef t(\frac{\sqrt{x}-1}{\sqrt{x}}\right) =$ $\frac12+\frac12e\log(x)-x\log\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right) = \frac12+\frac12e\log(x)-x\left(\sum_{n=1}^\infty \frac{1}{nx^{n/2}}\right)$
We need to be careful about $x=1$ in the line above. This is just one value though, so check it manually and assume $x>1$.

$\log\left(\sqrt{ex^e}\left(1 - \frac{1}{\sqrt{x}} \right)^x\right) < \frac12+\frac12e\log(x)-x\left(\frac{1}{\sqrt{x}}+\frac{1}{2x}+\frac{1}{3\ sqrt{x}^3}\right)$ $= \frac12e\log(x)-\sqrt{x}-\frac{1}{3\sqrt{x}} = f(x)$

Now let's find the maximum of $f(x)$.

$f'(x)=\frac{-3x+3e\sqrt{x}+1}{6x^{2/3}}$ so solving $f'(x)=0$ we see $x=\frac16\left(2+3e^2+e\sqrt{3(4+3 e^2)}\right)=x_0$.

But a little work shows $f(x_0)<0$ which proves the claim.

p.s. I'm sure there's a much more elegant proof.