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I need a function $f(x)$ that satisfies the properties bellow for all integers $k$

$$ \frac{\log(k+1)}{k+1}-\log\left(1+\frac 1 k\right)+f(k+1)-f(k)<0 \ $$ $$ \lim_{k \rightarrow \infty} f(k)=0 $$

I don't think it should be very hard sense if I let $f(x)=0$, the entire thing is already very close to zero. In addition if you find a function that doesn't work for the first few values 1,2,3.. etc, thats fine too. I would appreciate any help, thanks.

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You shouldn't use $x$ for variables ranging over integers, that can be rather confusing. I changed some $x$ to $k$, I hope you don't mind. –  tomasz Nov 13 '12 at 23:31
    
I assume you mean the inequality holds for all positive integers $k$? Otherwise $k = 0$ or $k = -1$ will lead to undefined terms. –  Benjamin Dickman Nov 14 '12 at 0:15
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1 Answer

The $f$ you seek cannot exist.

Define $f(k+1) - f(k) \triangleq -\varepsilon_{k+1}$. Since

$$ f(k) = f(1) - \sum_{j=2}^{k} \varepsilon_{j}, $$

if we assume that

$$ \varepsilon_{k+1} > \frac{\log(k+1)}{k+1}-\log\left(1+\frac 1 k\right) \sim \frac{\log k}{k} $$

for $k$ large enough, we would necessarily have $f(k) \to -\infty$ as $k \to \infty$ by the comparison test.

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