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Let $f^{k+1}(x)=f(f^k(x))$ and $f^0(x)=f(x)>x$ such that $\displaystyle{\frac{df}{dx}(x)>0}$ for all $x>0$. Does it imply that $$g(x)=\sum_{k=1}^\infty \frac{1}{f^k(x)}<\infty$$ for all $x>0$?

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Usually one takes $f^0(x)=x$ and $f^1(x)=f(x)$. – Michael Hardy Nov 16 '11 at 16:43
up vote 6 down vote accepted

No. Here’s an even easier example. Take $f(x)=x+1$; then $f^k(x)=x+k+1>x$, and $f\;'(x)=1$ for all $x$. But $$\sum_{k\ge 1}\frac1{f^k(x)} = \sum_{k\ge 2}\frac1{x+k},$$ which clearly diverges for $x>0$.

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No. Consider $f(x)=x+c/(x+1)$ with $0<c\leqslant1$. Then $f'(x)\gt1-c\geqslant0$ for every $x\gt0$ and $f(x)^2=x^2+2c+o(1)$ when $x\to+\infty$ hence, for every fixed $x\gt0$, $[f^k(x)]^2=2ck+o(k)$ when $k\to\infty$. In particular, $1/f^k(x)$ is of order $1/\sqrt{k}$ hence $g(x)$ is infinite for every $x\gt0$.

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No. If there is no typo in the question, take $f(x) = x$ (then $f^k(x) = x$ independently of $k$).

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Sorry for typo. its f(x)>x – 100101011011010011010110101110 Nov 16 '11 at 16:07

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