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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. – Did Jan 23 '12 at 1:27
@DidierPiau ? 2^(n+1)>2^n+...2+1 – Upcoming Ms.USA 2012 Jan 23 '12 at 2:13
Precisely. Hence $f(n+1)\lt f(n)+\cdots+f(1)$ fails although $f(n)\lt ab^n$ holds for $a=b=2$. – Did Jan 23 '12 at 2:18
@DidierPiau a=b=2 is not every a>0 and b>1 – Upcoming Ms.USA 2012 Jan 23 '12 at 2:24
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. – Did Jan 23 '12 at 6:45
up vote 3 down vote accepted

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 $f(2) <f(1)$...

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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.

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