[edit:] yeah yeah, I misunderstand it... DON'T READ THIS!!! For $ n\leq 2008$ take $ a_i = 1$. When $ 2008k < n \leq 2008(k + 1)$ take $ a_i = k + 1$. Then $ \sum a_i = n(k + 1) < n^2$ and $ \sum \frac 1{a_i} = \frac n{k + 1} \leq 2008$. I probably made some mistake bye

SpongeBob wrote: For $ n\leq 2008$ take $ a_i = 1$. I probably made some mistake It's already gone if you start with this (you've probably misread the question). It seems that the answer is no, there is no such sequence $ a_n$ (such that the sequence $ h_n = \sum_{k=1}^n \frac 1{a_k}$ is bounded), but I don't have a clear proof yet.

Is the inequality $ \sum_{i=n+1}^{2n}{\frac{1}{a_i}}\geq 1/4$ clear? Then the solution is clear, too. Ooops, I see I'm posting in the inequalities forum. Sorry....

All right, let's just write it out loud then: $ a_{k+1}+\cdots + a_{2k} < a_1+\cdots + a_{2k} \leq 4k^2$, therefore \[ \frac k{\cfrac 1{a_{k+1}}+\cdots+\cfrac 1{a_{2k}}} \leq \frac {a_{k+1} +\cdots + a_{2k}}{k} < 4k ,\] by AM-GM, which means that $ \frac 1{a_{k+1}} + \cdots + \frac 1{a_{2k}} > \frac 14$. Obviously this implies that the sequence $ s_n= \sum_{k=1}^n \frac 1{a_k}$ is unbounded, so there is no such sequence $ a_n$.

I was actually trying to prove something much harder, and as a matter of fact I'm not even sure that's true: that if $ b_i$ is a sequence such that $ n^2 \geq a_1+\cdots +a_i \geq b_1+\cdots + b_i$ for all $ i$, then $ \sum_{i=1}^n \frac 1{a_i} \leq \sum_{i=1}^n \frac 1{b_i}$ with Abel sums and such, though it seems not to be true ...

I remember that sum ( k/ (a1+...+ak) ) < 2. (1/a1 +.1/a2 +... ) and 2 is the best constant on the right side

I prove that : $ \sum_{i = 1}^n \frac {1}{a_{i}} \geq \sum_{i = 1}^n \frac {1}{2i - 1}$ by Cauchy Inequality .

harazi wrote: I see I'm posting in the inequalities forum. It's very well!

khashi70 wrote: I prove that : $ \sum_{i = 1}^n \frac {1}{a_{i}} \geq \sum_{i = 1}^n \frac {1}{2i - 1}$ by Cauchy Inequality . I would like to see that proof, although I doubt very much it's correct.

Ok . Here is my proof : by Cauchy inequality we have : $ (\sum _{i = 1}^n \frac {1}{a_{i}})(\sum_{i = 1}^n \frac {a_{i}}{(2i - 1)^2})\geq (\sum _{i = 1}^n \frac {1}{2i - 1})^2$ now we need to show that $ \sum_{i = 1}^n \frac {a_{i}}{(2i - 1)^2}\leq \sum _{i = 1}^n \frac {1}{2i - 1}$. now we define $ S_{k} = \sum_{i = 1}^k a_{i}$ , so we have $ S_{k}\leq k^{2}$ : $ (\frac {1}{(2n - 1)^2})S_{n}\leq \frac {n^{2}}{(2n - 1)^{2}}$ and for all $ 1\leq i\leq n - 1$ we write : $ (\frac {1}{(2i - 1)^{2}} - \frac {1}{(2i + 1)^{2}})S_{i}\leq (\frac {1}{(2i - 1)^{2}} - \frac {1}{(2i + 1)^{2}}) i^{2}$ and adding these inequalities : $ \sum_{i = 1}^n \frac {a_{i}}{(2i - 1)^2} \leq \sum_{i = 1}^{n - 1} \frac {8i^{3}}{(4i^{2} - 1)^{2}} + \frac {n^{2}}{(2n - 1)^{2}}$ and now by induction u can show that : $ RH = \sum _{i = 1}^n \frac {1}{2i - 1}$

Valentin Vornicu wrote: I was actually trying to prove something much harder, and as a matter of fact I'm not even sure that's true: that if $ b_i$ is a sequence such that $ n^2 \geq a_1 + \cdots + a_i \geq b_1 + \cdots + b_i$ for all $ i$, then $ \sum_{i = 1}^n \frac 1{a_i} \leq \sum_{i = 1}^n \frac 1{b_i}$ with Abel sums and such, though it seems not to be true ... If the "majorizing" sequence is increasing, then it's true. See here: http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=A408&l=en

d.edgar wrote: Valentin Vornicu wrote: I was actually trying to prove something much harder, and as a matter of fact I'm not even sure that's true: that if $ b_i$ is a sequence such that $ n^2 \geq a_1 + \cdots + a_i \geq b_1 + \cdots + b_i$ for all $ i$, then $ \sum_{i = 1}^n \frac 1{a_i} \leq \sum_{i = 1}^n \frac 1{b_i}$ with Abel sums and such, though it seems not to be true ... If the "majorizing" sequence is increasing, then it's true. See here: http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=A408&l=en Actually if we have that $(a_1,...,a_n) \succ (b_1,...,b_n)$ then applying Karamata inequality to the concave function $f(x)=\frac{1}{x}$ we get $ \sum_{i = 1}^n \frac 1{a_i} \leq \sum_{i = 1}^n \frac 1{b_i}$. But if we don't have the majorizing condition, I don't quite think that it is true Look at this in this way.... We can take the sequences such that $(b_1,...,b_n) \succ (a_1,...,a_n)$, then we have the same inequality, but with the reversed sign !

Can someone provide a full solution?

Official Solution The answer is no We will show that if $\sum_{i=1}^{n}{a_i} \leq n^2$ for any $n$ then $\sum_{i=1}^{2^n}{\frac{1}{a_i}} >\frac{n}{4}$ Since $\sum_{i=2^k+1}^{2^{k+1}}{a_i} <\sum_{i=1}^{2^{k+1}}{a_i} \leq 2^{2k+2}$, its follow that $\sum_{i=2^k+1}^{2^{k+1}}{\frac{1}{a_i}}>\frac{1}{4}$ And hence $\sum_{i=2}^{2^n}{\frac{1}{a_i}} =\sum_{k=0}^{n-1}{(\sum_{i=2^k+1}^{2^{k+1}}{\frac{1}{a_i}})} >\frac{n}{4}$ Finally $\sum_{i=1}^{2^n}{\frac{1}{a_i}} >\sum_{i=2}^{2^n}{\frac{1}{a_i}} >\frac{n}{4}$

I don't understand how the solution concludes to the desired result.

@above it shows that the series that should not be greater than $2008$ is actually unbounded, then implying that we can find an $n$ such that the series is strictly greater than $2008$, contradiction.

Let us prove that for any infinite sequence of positive reals satisfying $a_1+a_2+\dots+a_n \le n^2$, and each positive integer $k$ there is an $N\in\mathbb{N}$ for which $\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_N}>\frac{1}{2}k$. We will induct on $k$. $k=1$ is trivial, as $a_1\le1 \Rightarrow \frac{1}{a_1}\ge1>\frac{1}{2}$. Assume the induction hypothesis for $k$: $$\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_N}>\frac{1}{2}k$$Now to prove it for $k+1$, assume on the contrary that for any $m\in\mathbb{N}$, $\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_m}<\frac{1}{2}(k+1)$. Then, $$\frac{1}{2}k<\frac{1}{a_1}+\dots+\frac{1}{a_{N}}<\frac{1}{a_1}+\dots+\frac{1}{a_{N+m}}<\frac{1}{2}(k+1)$$from where it follows that $$\frac{1}{a_N}+\dots+\frac{1}{a_{N+m}}<\frac{1}{2}$$for any $m$. But we know from Cauchy that $(\frac{1}{a_N}+\dots+\frac{1}{a_{N+m}})(a_N+\dots+a_{N+m})\ge (m+1)^2>m^2$, which, combined with the above inequality, gives $a_N+\dots+a_{N+m}>2m^2$. So, by the problem condition, we get $(N+m)^2\ge a_1+\dots+a_{N+m}>2m^2$ for any positive integer $m$. But this means $N^2>m(m-2N)$ for any m, which clearly does not hold for sufficiently large $m$. Contradiction. The induction is completed, so we have shown that $\sum \frac{1}{a_i}$ is unbounded. NOTE: the choice of $\frac{1}{2}$ was arbitrary, and it can in fact be replaced by any real number $0<\lambda<1$, with all the steps remaining analogous.

I will post solution that uses mixing variables and proves stronger fact: $$ \displaystyle\sum_{i=1}^{c+1} \frac{1}{a_i} > 2008 $$where $ c = 2^{4014} $. Assume opposite for the sake of contradiction, that this statement is wrong. We define parallel sequence $ B = \{ b_1, b_2, \ldots, b_c \} $ such that $ \{ b_1, b_2, \ldots, b_c \} = \{ a_1, a_2, \ldots, a_c \} $ and $ b_1 \le b_2 \le \ldots \le b_c $. It is obvious that the condition: $ b_1+b_2+\ldots+b_n \le n^2$ is satisfied (we call that $ condition \,1 $) and also that this statement is true: $ \displaystyle\sum_{i=1}^{c} \frac{1}{a_i} \ge \displaystyle\sum_{i=1}^{c} \frac{1}{b_i}$. We can say that $ b_1+b_2+\ldots+b_c = c^2$, because this sum: $ \displaystyle\sum_{i=1}^{n} \frac{1}{b_i} $ decreases when we increase $ b_c $. Lemma: $ \frac{1}{x} + \frac{1}{y} \ge \frac{1}{x + d} + \frac{1}{y - d}, x,y > 0, x \le y $ and $ 0 \le d \le \frac{y-x}{2} $. Proof: $ \frac{1}{x} + \frac{1}{y} \ge \frac{1}{x + d} + \frac{1}{y - d} \Leftrightarrow \frac{x+y}{xy} \ge \frac{x+y}{(x + d)(y - d)} \Leftrightarrow xy \le (x + d)(y - d) \Leftrightarrow d^2 \le d(y - x) \Leftrightarrow d \le y - x $ which is true. Transformation: $ ({b_t}_i, {b_t}_j) \longmapsto ({b_{t+1}}_i, {b_{t+1}}_j) , {b_{t+1}}_i = {b_t}_i + d, {b_{t+1}}_j = {b_t}_j - d $, where $ d $ is maximum integer such that: $ {b_{t+1}}_i \le {b_{t+1}}_j $. We also apply another condition on choosing $d$: If $ \displaystyle\sum_{i=1}^{i} {b_t}_i \le i^2 $, then $ d \le i^2 - \displaystyle\sum_{i=1}^{i} {b_t}_i $, else $ d = 0 $. We use this transformation on pair $ (i, j), 1 \le i < j \le c $, so $ {b_t}_i \le {b_t}_j $. Because of Lemma: $ \frac{1}{{b_t}_i} + \frac{1}{{b_t}_j} \ge \frac{1}{{b_{t+1}}_i} + \frac{1}{{b_{t+1}}_j} $. If we use transformations on the these pairs in this order: $ \{ (1,2), (1,3), \ldots, (1,c), (2,3), \ldots, (2,c), \ldots, (c-1,c) \} $ (there are $ c-1 $ trasformation made on each term), we will get sequence: $ B_c= \{ {b_c}_1, {b_c}_2, \ldots, {b_c}_c \} $ (if we start from sequence $ B_1 $). I will prove that this equality is true: $ b_n = 2n - 1 $. Assume, for the sake of contradiction, that there exist $ k \in \{2, 3, \ldots, c \} $ such that $ {b_c}_k > 2k - 1 $ and $ {b_c}_i \le 2k - 1 $ for each $ i \in \{1, 2, \ldots, k-1 \} $ ($ k \neq 1 $ because we always choose $ d $ such that $ {b_{t+1}}_1 \le 1 $). After each transformation, $ condition \,1 $ is satisfied, so: $ b_1+b_2+\ldots+b_{k-1} < (k-1)^2$, but that is in contradiction with choosing maximum $ d $ in pairs: $ \{ (1,k), (1,k), \ldots, (k-1,k) \} $. Contradiction, $ b_n = 2n - 1 $, for each $ n \in \{1, 2, \ldots, c \} $. We are now done because: $$ \displaystyle\sum_{i=1}^{c+1} \frac{1}{a_i} > \displaystyle\sum_{i=1}^{2^{4014}} \frac{1}{a_i} \ge \displaystyle\sum_{i=1}^{2^{4014}} \frac{1}{b_i} \ge \displaystyle\sum_{i=1}^{2^{4014}} \frac{1}{{b_c}_i} = \displaystyle\sum_{i=1}^{2^{4014}} \frac{1}{i} = \frac{1}{1} + \displaystyle\sum_{i=1}^{4014} \displaystyle\sum_{j=1}^{2^{i-1}} \frac{1}{2^{i-1} + j} \ge \displaystyle\sum_{i=1}^{4014} \displaystyle\sum_{j=1}^{2^{i-1}} \frac{1}{2^{i}} = 1 + \displaystyle\sum_{i=1}^{4014} \frac{1}{2} = 1 + \frac {4014}{2} = 2008 $$ This is contradiction, so answer is no, there is no such a sequence.

Looking at Romanian mathematical competitions 2002 booklet, I just found that the same problem was already posted at "Sixth regional contest 'Alexandru Papiu Ilarian' " that was organized in 2001. There, part b) of the first problem for grade 12 was to prove that under the condition that for positive reals $x_{1}$, $x_{2}$, $\dots$ inequality $x_{1} + x_{2} + \dots x_{n} \leq n ^ {2}$ holds for all positive integers $n$ then we have \[ \lim_{n\rightarrow \infty} \left(\frac{1}{x_{1}} + \frac{1}{x_{2}} + \dots + \frac{1}{x_{n}} \right) = \infty \]so the problem seems to be completely non-original (unless I am completely wrong). Clearly, I do not claim that the author whom this problem was attributed (as far as I remember it was Nikolai Nikolov from Bulgaria) at Balkan M.O. copied it from the other resources. Although the official solutions at both places are pretty much the same, due to the nature of the problem, it is quite likely that two authors might have come up with it independently. However, it is quite interesting that no one in the jury and problem selection committee noticed this.

No such sequence exists. The key is to observe for each \(k\) that \[\sum_{i=k+1}^{2k}\frac1{a_i}\ge k\cdot\frac1{(a_{k+1}+\cdots+a_{2k})/k}\ge\frac k{4k^2/k}=\frac14.\]It follows that \[\sum_{i=1}^{2^{8032}}\frac1{a_i}\ge a_1+{\underbrace{\frac14+\cdots+\frac14}_{8032}}>2008.\]