Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
$f(n)=2^n$ 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
1  
@DidierPiau a=b=2 is not every a>0 and b>1 –  Upcoming Ms.USA 2012 Jan 23 '12 at 2:24
2  
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
show 3 more comments

2 Answers

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

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.