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?
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?
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)
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.