If f(0) > 0 the limit of a_n is 0. If f(0) < 0 the limit is infinity. If f(0) = 0 then it depends if f is positive, negative or keeps changing sign at right of 0 and how fast is decreasing toward 0.

You can write a_n as (1 / (1 + S/n))^n where by S I denoted f(1/2) + ... f(1/n). By Stoltz, S/n has limit f(0). This implies what I said above.

Gabriel, please think about the cases f(0)=-2 and f(0)

Dear Gabriel, I think about the limit, when f(0)=0.

With your notation, the limit in this case is e^lim(-nS/(n+S)). What condition on f to obtain a finite lim(-nS/(n+S))?

Just an example, possibly it helps. I will not be surprised it it is known:-)

If f is continuous, positive, increasing and integrable wrt to the weight 1/x^2 in [0;1/2],

then the limit of a_n is equal to exp( - \sum_{j=2}^{\infty} f(1/j), which is finite.

I hope there is no error:-)

Thanks Joachim! I think your example has no error, but the conditions on f are very restrictive.

Meanwhile, I found the following: If f satisfies f(x)>=x for all x in [0,1/2], then S=f(1/2)+...+f(1/n)>=1/2+...+1/n, so lim[f(1/2)+...+f(1/n)] is infinity.

lim(n/(n+S)) =1 when f(0)=0. Consequently lim(nS/(n+S)) is infinity. Therefore limit of a_n is 0 in case f(0)=0, if f(x)>=x. I hope too, there is no error !!

Hi Dinu, Yes your case is also ok, I think.

However, let me soften my condition by edding that monotonicity can be replaced by finite variation and integrability of both parts in the decomposition into the incresing and decresing parts, of course:-)

Currently, no further extension is obtained by me.

Dinu Teodorescu

That's me again. Let me share the following opinion ABOUT the type of the problem, if someone asks for fimite limit of a_n:

OBVIOUSLY, the necessary and sufficient condition sounds:

(*) the series of f(1/n) converges conditionally to a finite number.

Thus the question is which tools are you admitting for characterizing continuous functions with respect to this condition. In other words: which classes of functions you mean. Note also, please, that even continuity is not substantial (though the convergence to 0 of the sequence of values at the given points remains NECESSARY).

Due to this the whole class of functions satisfying (*) contains very strange functions, even non-measurable in any form.

Therefore let me close this brief explanation by a continuous function causing than a_n approches 0:

f(x) = F(x) sin(\pi/x), with continuous F such that F(0)=0.

Dear Joachim, The problem was proposed by someone from University of Nis in SEEMOUS Math Competition, organized by Mathematical Society of SouthEastern Europe for Math University Students. For reasons I don't know, the problem was not selected to be given effectively in the contest(perhaps too easy or unclear statement and requirements.....or too difficult).

A friend from Tech University of Cluj( known as Klausenburg in Deutschland !!) sent me the problem. As proposed, the problem has two more requirements:

ii) Give an example of continuous function with f(0)=0, so that (a_n) converges.

iii) Give an example of continuous function with f(0)=-2, so that (a_n) do not converges.

Requirement to search conditions on f, is my requirement! I do not understand why the author of problem require continuity on R. I thought that the author has given a nice solution using continuity...perhaps!

Clearly we can replace continuity with other conditions, so that series of f(1/n) converges.

Dear Dinu Teodorescu

Thus everything is clear, JoaD

PS. I am happy that I have given wider class of functions. As an example one can take f(x) = x^{1+\epsilon} , x >=0;with arbitrary positive \epsilon. Really, ii is a very simple problem. Problem iii is left for other followers.

Best regards, Joachim Domsta

Dear Joachim,

Thanks for your interest on this problem and for your interesting posts!

Best regards, Dinu Teodorescu

Dear Dinu, you are welcome, and I thank you for a problem I was able to solve:-)

Best regards, Joachim Domsta