Let $ S_n = \sum_{i = 1}^n a_i \forall n \ge 0$. We have $ r(S_n - S_{n - 1}) \ge S_{n + 1}$.

Let us suppose $ S_{n + 1} \ge kS_n \forall n \in \mathbb{Z}_ +$ and some real $ k$. We have $ r(S_n - S_{n - 1}) \ge S_{n+1} \ge kS_{n} \Rightarrow S_n(r - k) \ge rS_{n - 1}$ $ S_{n - 1} > 0$ gives $ r - k > 0$ and $ S_n \ge \frac {r}{r - k}S_{n - 1}$. In other words, $ S_{n + 1} \ge \frac {r}{r - k}S_n \forall n \in \mathbb{Z}_ +$. Now, as we know that $ S_{n + 1} > S_n$, if we define $ a_0 = 1$ and $ a_{n + 1} = \frac {r}{r - a_n}$, we see that this sequence is well defined and all its terms are $ > 0$. Suppose $ r < 4$. Now, we can prove that this sequence is increasing. In fact, $ \frac {r}{r - a_n} > a_n \Longleftrightarrow r(a_n - 1) < a_n^2$ and $ r(x - 1) < 4(x - 1) \le x^2 \Longleftrightarrow (x - 2)^2 \ge 0$. But $ a_n \le \frac {S_2}{S_1}$, therefore $ a_n$ is bounded. Since this sequence is increasing, it is convergent to some limit $ L$. Now it is easy! $ L = \frac {r}{r - L} \Rightarrow L^2 - Lr + r = 0$ We must have $ 0 \le \Delta = r^2 - 4r$.

why you take $ a_{n + 1} = \frac {r}{r - a_n}$ and please explain this $ a_n \le \frac {S_2}{S_1}$

feliz wrote: ... if we define $ a_0 = 1$ and $ a_{n + 1} = \frac {r}{r - a_n}$, we see that this sequence is well defined and all its terms are $ > 0$. ... This is false in general. If we consider the function $f(x) = \dfrac {r} {r-x}$ and we consider its iterates at $x=1$ we may run into trouble, especially when $r<4$; just see that for $r=2$ or $r=3$ we reach a value equal to $r$, and so we cannot continue. The claim $a_n \leq \dfrac {S_2} {S_1}$ is also unjustified (and other numerous typos and errors I have corrected). However, the idea is good. As there, define $S_n = \sum_{i=1}^n a_i$, so the sequence $(S_n)_{n\geq 1}$ is increasing, forcing $r>1$. Now take $K = \{k \in \mathbb{R}_+^* \mid S_{n+1} \geq kS_{n} \textrm{ for all } n \in \mathbb{Z}_+^*\}$, and $\ell = \sup K$. This supremum exists, since $1\in K$ and $K$ is upper bounded by $\dfrac {S_2} {S_1}$; moreover $\ell\geq 1$, and $\ell \in K$. But now $\ell S_{n+1} \leq S_{n+2} \leq r(S_{n+1} - S{n})$ is equivalent to $(r-\ell)S_{n+1} \geq rS_n$, forcing $\ell < r$. Therefore $S_{n+1} \geq \dfrac {r} {r-\ell} S_n$, and this for all $n$, so $\dfrac {r} {r-\ell} \in K$, therefore $\dfrac {r} {r-\ell} \leq \ell$. This writes as $\ell^2 - r\ell + r \leq 0$, so we must have for its discriminant $\Delta = r^2 - 4r \geq 0$, thus $r\geq 4$. Let us finally notice that $r=4$ is indeed reachable, by taking, for any positive real number $a$, the values $a_1=a$, and $a_n=2^{n-2}a$ for $n\geq 2$; then $S_{n+1} = 2S_n$, so $S_{n} = 2^{n-1}a$, $\ell = 2$, $r=4$ and equalities allover.

Can someone post full and correct solution???

I didn't understand anything

Any solutions please

Can someone explain this part in mavropnevma's solution: mavropnevma wrote: Now take $K = \{k \in \mathbb{R}_+^* \mid S_{n+1} \geq kS_{n} \textrm{ for all } n \in \mathbb{Z}_+^*\}$, and $\ell = \sup K$. This supremum exists, since $1\in K$ and $K$ is upper bounded by $\dfrac {S_2} {S_1}$; moreover $\ell\geq 1$, and $\ell \in K$.

Here is what I believe to be a very natural and straightforward solution. Clearly $\frac{a_{n+1}}{a_n} \le r$ is bounded. So let $L=\limsup \frac{a_{n+1}}{a_n}$ (this sounds fancy but is really just the smallest essential upper bound on the ratios, i.e. upon ignoring the first finitely many values). Now for any fixed $L_1<L<L_2$, we have $\frac{a_{n+1}}{a_n}<L_2$ for all sufficiently large $n$ and hence for any fixed $k$, we have for all large $n$, that \[a_n+a_{n-1}+a_{n-2}+\dots+a_{n-k} \ge \left(1+\frac{1}{L_2}+\frac{1}{L_2^2}+\dots+\frac{1}{L_2^k}\right) a_n.\]Moreover, for infinitely many $n$ we have $a_{n+1}>L_1a_n$ and hence \[r \ge L_1+1+\frac{1}{L_2}+\frac{1}{L_2^2}+\dots+\frac{1}{L_2^k}.\]Now the rest is straightforward: Since $L_1<L<L_2$ were arbitrary, we get that \[r \ge L+1+\frac{1}{L}+\frac{1}{L^2}+\dots+\frac{1}{L^k}.\]Moreover, as $k$ was arbitrary, we get \[r \ge L+1+\frac{1}{L}+\frac{1}{L^2}+\dots=L+1+\frac{1}{L-1}=2+(L-1)+\frac{1}{L-1} \ge 4\]as desired (note that $L>1$ is already implied from the first inequality by the fact that $r$ is finite).

Let $S_n=\sum_{i=1}^{n}a_i$ then $S_{n+1}>S_n.$ We have: $S_{n+1} \leq r\left ( S_n-S_{n-1} \right ) \, \rightarrow \frac{S_{n-1}}{S_n}+\frac{1}{r.\frac{S_n}{S_{n+1}}} \leq 1$ Define $x_n=\frac{S_{n-1}}{S_n}$ then $1 \geq x_n+\frac{1}{r.x_{n+1}}, \, (1) $ Assume $r<4$, then: $1\geq x_n + \frac{1}{4.x_{n+1}} \geq 2\sqrt{\frac{x_n}{4.x_{n+1}}} \rightarrow x_n \leq x_{n+1} \leq ... \leq 1$ So $\textrm{lim}x_n=L \leq 1.$ From $(1)$, we know: $1 \geq L+\frac{1}{r.L} \rightarrow r \geq \frac{1}{L.(1-L)}\geq 4,$ adsurb.

We can prove this by contradiction. Suppose that r < 4. We will show that this leads to a contradiction. Since r < 4, we have a_1 + a_2 + … + a_{m+1} <= (r)a_m < 4a_m for each positive integer m. If we let S_m = a_1 + a_2 + … + a_m, we can write: S_{m+1} = S_m + a_{m+1} < 4a_m + a_{m+1} = 4(a_1 + a_2 + … + a_m) + a_{m+1} = 4S_m + a_{m+1} Subtracting 4S_m from both sides, we get: S_{m+1} - 4S_m < a_{m+1} This simplifies to: (1-4)S_m < a_{m+1} Since S_m is the sum of the first m terms of the sequence, it is a positive real number. Therefore, the left-hand side of the inequality is negative, which means that the inequality has no solutions. This is a contradiction, so our assumption that r < 4 must be false. Therefore, we must have r >= 4. I hope this helps! Let me know if you have any questions.