Is there a continuous and monotone function $1>f(n)>0$, such that $\lim_{n\to +\infty}(f(n)\ln(n+1)-\ln(n))=1\mbox{ and }f(n)>\frac{\ln(n)}{\ln(n+1)}?$
$f(n)$ such that $\lim_{n\to +\infty}(f(n)\ln(n+1)-\ln(n))=1$
1
$\begingroup$
real-analysis
-
1$f(n) \ln (n+1) - \ln n \leq \ln (n+1) - \ln n \to 0$ – 2011-12-07
1 Answers
1
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