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How to find or bound the fastest growing function $f(n)>0$, with $f'(n)>0$ such that $f(n+1)<\sum_{j=1}^n f(j)$ ?

Is $\ln(f(n))\ll n$ necessary and sufficient?

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    $f(n)=2^n$ fails the test.2012-01-23
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    @DidierPiau ? 2^(n+1)>2^n+...2+12012-01-23
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    Precisely. Hence $f(n+1)\lt f(n)+\cdots+f(1)$ fails although $f(n)\lt ab^n$ holds for $a=b=2$.2012-01-23
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    @DidierPiau a=b=2 is not every a>0 and b>12012-01-23
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    Right. You might wish to reformulate the (at the moment, quite ambiguous) condition you have in mind. Is it equivalent to $\log f(n)\ll n$?2012-01-23
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    Upcoming, What do you think is meant by the phrase $\log f(n) < < n?$2012-01-23
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    @WillJagy Not sure, log(f(n))0 and sufficiently large n?2012-01-23
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    Although the OP did not see fit to mention it, there is not much in common between the current version of the question and the original one, which the 5 first comments above address. Odd as well is to accept an answer which (competently) deals with the necessary condition but not the sufficient one, without waiting for some more complete contributions.2012-01-23

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Let $f(1)=a$. Then $f(2) < a$ and

$$f(3) < f(1)+f(2)=2a \,.$$

$$f(4) < f(1)+f(2)+f(3)< a+a+2a =4a \,.$$

By induction you can prove now that

$$f(n)< 2^{n-2} a \,.$$

Which seems to be exactly the type of result you seek.

P.S. The conditions $f(2)< f(1)$ seems odd, especially since you want a positive first derivative... Also, I think that any function of the type $f(x)=a2^x-\epsilon$ satisfies the requirements, excepting for $f(2) ...

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You asked about necessary and sufficient. $$ f(n) = 3^n $$ satisfies $$ \log f(n) = n \log 3 $$ but $$ f(n+1) > \sum_{j=1}^n f(j) $$ So the $\log$ condition you name is not sufficent.