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Math Help - An upper bound for a series

  1. #1
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    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}}.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by NonCommAlg View Post
    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.
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