# $f(n)$ such that $\lim_{n\to +\infty}(f(n)\ln(n+1)-\ln(n))=1$

Is there a continuous and monotone function $1>f(n)>0$, such that $$\lim_{n\to +\infty}(f(n)\ln(n+1)-\ln(n))=1\mbox{ and }f(n)>\frac{\ln(n)}{\ln(n+1)}?$$

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$f(n) \ln (n+1) - \ln n \leq \ln (n+1) - \ln n \to 0$ – sdcvvc Dec 7 '11 at 16:34

If so, then $\ln{(n+1)^{f(n)}\over n}$ would converge to 1. But then, ${(n+1)^{f(n)}\over n}$ would have to converge to $e$. But, since $0<f(n)<1$, we have ${(n+1)^{f(n)}\over n}<{n+1\over n}\rightarrow1$.