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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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    I assume you mean the inequality holds for all *positive* integers $k$? Otherwise $k = 0$ or $k = -1$ will lead to undefined terms.2012-11-14

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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.